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# How do you divide (-x^3 - x^2-6x+5 )/((-x + 10 )?
##### 1 Answer
Feb 21, 2016
(−x^3−x^2−6x+5)/(-x+10) is
${x}^{2} + 11 x + 116 - \frac{1155}{- x + 10}$
#### Explanation:
To divide −x^3−x^2−6x+5 by $- x + 10$, first we observe that $- x$ goes $\frac{- {x}^{3}}{-} x = {x}^{2}$ times in $- {x}^{3}$.
Hence, x^2(-x+10)-10x^2−x^2−6x+5 or
(note we have added and subtracted 10x^2) x^2(-x+10)-11x^2−6x+5
As $- 11 {x}^{2} / - x = 11 x$, above can be written is
x^2(-x+10)+11x(-x+10)-110x−6x+5 or
${x}^{2} \left(- x + 10\right) + 11 x \left(- x + 10\right) - 116 x + 5$
and as $- 116 \frac{x}{-} x = 116$, above can be written as
${x}^{2} \left(- x + 10\right) + 11 x \left(- x + 10\right) + 116 x \left(- x + 10\right) - 1160 + 5$ or
${x}^{2} \left(- x + 10\right) + 11 x \left(- x + 10\right) + 116 x \left(- x + 10\right) - 1155$ or
$\left({x}^{2} + 11 x + 116\right) \left(- x + 10\right) - 1155$ or
Hence (−x^3−x^2−6x+5)/(-x+10)# is
${x}^{2} + 11 x + 116 - \frac{1155}{- x + 10}$ |
# Dividing Whole Numbers with One-Digit Divisors
Dividing Whole Numbers with One-Digit Divisors
1 / 18
Slide 1: Tekstslide
MathPrimary EducationAge 9
In deze les zitten 18 slides, met interactieve quizzen en tekstslides.
Lesduur is: 45 min
## Onderdelen in deze les
Dividing Whole Numbers with One-Digit Divisors
#### Slide 1 -Tekstslide
Deze slide heeft geen instructies
Learning Objective
Understand and apply strategies based on place value, properties of operations, and the relationship between multiplication and division to find whole-number quotients and remainders with one-digit divisors and up to four-digit dividends.
#### Slide 2 -Tekstslide
Deze slide heeft geen instructies
What do you already know about dividing whole numbers?
#### Slide 3 -Woordweb
Deze slide heeft geen instructies
Review of Place Value
Place value is the value of a digit based on its position in a number. For example, in the number 3,245, the 2 is in the hundreds place, so its value is 200.
#### Slide 4 -Tekstslide
Deze slide heeft geen instructies
Properties of Operations
Review the commutative and associative properties of multiplication and division and how they can help in solving division problems.
#### Slide 5 -Tekstslide
Deze slide heeft geen instructies
Multiplication and Division Relationship
Discuss how multiplication and division are related, and how understanding one operation can help in solving problems involving the other.
#### Slide 6 -Tekstslide
Deze slide heeft geen instructies
Division with One-Digit Divisors
Learn and practice division with one-digit divisors using strategies based on place value and the relationship between multiplication and division.
#### Slide 7 -Tekstslide
Deze slide heeft geen instructies
Using Remainders
Understand how to interpret and use remainders in division problems, and when it is appropriate to include the remainder in the quotient.
#### Slide 8 -Tekstslide
Deze slide heeft geen instructies
Practicing Division
Apply the strategies learned to solve division problems with one-digit divisors and up to four-digit dividends.
#### Slide 9 -Tekstslide
Deze slide heeft geen instructies
Word Problems
Solve word problems involving division using place value, properties of operations, and the relationship between multiplication and division.
#### Slide 10 -Tekstslide
Deze slide heeft geen instructies
Review strategies for checking division answers, such as using multiplication to verify the quotient and remainder.
#### Slide 11 -Tekstslide
Deze slide heeft geen instructies
Interactive Activity
Engage in a group activity where students create division word problems for their peers to solve, incorporating the strategies learned in the lesson.
#### Slide 12 -Tekstslide
Deze slide heeft geen instructies
Review and Recap
Summarize the key concepts learned in the lesson and address any questions or misunderstandings.
#### Slide 13 -Tekstslide
Deze slide heeft geen instructies
Assessment
Assess students' understanding of division with one-digit divisors through a short quiz or problem-solving activity.
#### Slide 14 -Tekstslide
Deze slide heeft geen instructies
Homework
Assign practice problems or a reflection activity for homework to reinforce the concepts covered in the lesson.
#### Slide 15 -Tekstslide
Deze slide heeft geen instructies
Write down 3 things you learned in this lesson.
#### Slide 16 -Open vraag
Have students enter three things they learned in this lesson. With this they can indicate their own learning efficiency of this lesson.
Write down 2 things you want to know more about.
#### Slide 17 -Open vraag
Here, students enter two things they would like to know more about. T
his not only increases involvement, but also gives them more ownership.
Ask 1 question about something you haven't quite understood yet.
#### Slide 18 -Open vraag
The students indicate here (in question form) with which part of the material they still have difficulty. For the teacher, this not only provides insight into the extent to which the students understand/master the material, but also a good starting point for the next lesson. |
# > Fractions
### Fractions
• A Fraction is a number , of the form where a and b are integers and b ≠ 0. The integer a is called the numerator of the fraction, and b is called the denominator. For e.g. is a fraction in which -7 is the numerator and 5 is the denominator. Such numbers are called rational numbers.
• If both the numerator a and denominator b are multiplied by the same nonzero integer, the resulting fraction will be equivalent to . For e.g.
• If both the numerator and denominator have a common factor, then the numerator and denominator can be factored and reduced to an equivalent fraction. For e.g.
• To add two fractions with the same denominator, you add the numerators and keep the same denominator. For e.g.
• To add two fractions with different denominators, first find a common denominator, which is a common multiple of the tow denominators. Then convert both fractions to equivalent fractions with the same denominator. Finally, add the numerators and keep the common denominator. For e.g. to add the fractions and , use the common denominator 15:
• The same method applies to subtraction of fractions.
• To multiply two fractions, multiply the two numerators and multiply the two denominators. For e.g.
• To divide one fraction by another, first invert the second fraction, that is, find its reciprocal, and then multiply the first fraction by the inverted fraction. For e.g.
• An expression such as is called mixed number. It consists of an integer part and a fraction part; the mixed number means . To convert a mixed number to an ordinary fraction, convert the integer part to an equivalent fraction and add it to the fraction part. For e.g.
• Note that numbers of the form , where either a or b is not an integer and b , are fractional expressions that can be manipulated just like fractions. For e.g. the numbers can be added together as follows. |
# 2012 AMC 12B Problems/Problem 14
## Problem
Bernardo and Silvia play the following game. An integer between $0$ and $999$ inclusive is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and passes the result to Silvia. Whenever Silvia receives a number, she addes $50$ to it and passes the result to Bernardo. The winner is the last person who produces a number less than $1000$. Let $N$ be the smallest initial number that results in a win for Bernardo. What is the sum of the digits of $N$?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$
## Solution
### Solution 1
The last number that Bernado says has to be between 950 and 999. Note that 1->2->52->104->154->308->358->716->776 contains 4 doubling actions. Thus, we have $x \rightarrow 2x \rightarrow 2x+50 \rightarrow 4x+100 \rightarrow 4x+150 \rightarrow 8x+300 \rightarrow 8x+350 \rightarrow 16x+700$.
Thus, $950<16x+700<1000$. Then, $16x>250 \implies x \geq 16$. If $x=16$, we have $16x+700=956$. Working backwards from 956,
$956 \rightarrow 478 \rightarrow 428 \rightarrow 214 \rightarrow 164 \rightarrow 82 \rightarrow 32 \rightarrow 16$.
So the starting number is 16, and our answer is $1+6=\boxed{7}$, which is A.
### Solution 2
Work backwards. The last number Bernardo produces must be in the range $[950,999]$. That means that before this, Silvia must produce a number in the range $[475,499]$. Before this, Bernardo must produce a number in the range $[425,449]$. Before this, Silvia must produce a number in the range $[213,224]$. Before this, Bernardo must produce a number in the range $[163,174]$. Before this, Silvia must produce a number in the range $[82,87]$. Before this, Bernardo must produce a number in the range $[32,37]$. Before this, Silvia must produce a number in the range $[16,18]$. Bernardo could not have added 50 to any number before this to obtain a number in the range $[16,18]$, hence the minimum $N$ is 16 with the sum of digits being $\boxed{7}$. |
# Determine the Laplace transform of the given function f. f(t)=(t -1)^2 u_2(t)
Question
Laplace transform
Determine the Laplace transform of the given function f.
$$f(t)=(t -1)^2 u_2(t)$$
2021-03-08
Step 1
Here laplace transform of given function is to be calculated.
Given function is $$f(t)=(t -1)^2 u_2(t)$$
So expand the function as follows,
$$f(t)=t^2u(t)+u(t)-2tu(t)$$
Take Laplace of the function f(t)
Step 2
Then,
$$L(f(t))=L(t^2u(t)+u(t)-2tu(t))$$
$$=\frac{2}{s^3}+\frac{1}{s}-\frac{2}{s^2}$$
Hence the Laplace Transform of the $$f(t)=(t-1)^2u(t) \text{ is } \frac{2}{s^3}+\frac{1}{s}-\frac{2}{s^2}$$
### Relevant Questions
determine a function f(t) that has the given Laplace transform
$$F(s)=\frac{4s+5}{s^{2}+9}$$
determine a function f(t) that has the given Laplace transform F(s).
$$F(s) = \frac{3}{(s^2)}$$
$$\text{Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by }$$
$$F(s)=\int_0^\infty e^{-st} f(t)dt \(\text{where we assume s is a positive real number. For example, to find the Laplace transform of } f(t)=e^{-t} \text{ , the following improper integral is evaluated using integration by parts:} \(F(s)=\int_0^\infty e^{-st}e^{-t}dt=\int_0^\infty e^{-(s+1)t}dt=\frac{1}{s+1}$$
$$\text{ Verify the following Laplace transforms, where u is a real number. }$$
$$f(t)=t \rightarrow F(s)=\frac{1}{s^2}$$
The function
$$\begin{cases}t & 0\leq t<1\\ e^t & t\geq1 \end{cases}$$
has the following Laplace transform,
$$L(f(t))=\int_0^1te^{-st}dt+\int_1^\infty e^{-(s+1)t}dt$$
True or False
Given the function $$\begin{cases}e^{-t}& \text{if } 0\leq t<2\\ 0&\text{if } 2\leq t\end{cases}$$
Express f(t) in terms of the shifted unit step function u(t -a)
F(t) - ?
Now find the Laplace transform F(s) of f(t)
F(s) - ?
Let f(t) be a function on $$\displaystyle{\left[{0},\infty\right)}$$. The Laplace transform of fis the function F defined by the integral $$\displaystyle{F}{\left({s}\right)}={\int_{{0}}^{\infty}}{e}^{{-{s}{t}}} f{{\left({t}\right)}}{\left.{d}{t}\right.}$$ . Use this definition to determine the Laplace transform of the following function.
$$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}-{t}&{0}<{t}<{1}\\{0}&{1}<{t}\end{matrix}\right.}$$
Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by
$$F(s)=\int_0^\infty e^{-st}f(t)dt$$
where we assume s is a positive real number. For example, to find the Laplace transform of f(t) = e^{-t}, the following improper integral is evaluated using integration by parts:
$$F(s)=\int_0^\infty e^{-st}e^{-t}dt=\int_0^\infty e^{-(s+1)t}dt=\frac{1}{(s+1)}$$
Verify the following Laplace transforms, where u is a real number.
$$f(t)=1 \rightarrow F(s)=\frac{1}{s}$$
Solution of I.V.P for harmonic oscillator with driving force is given by Inverse Laplace transform
$$y"+\omega^{2}y=\sin \gamma t , y(0)=0,y'(0)=0$$
1) $$y(t)=L^{-1}\bigg(\frac{\gamma}{(s^{2}+\omega^{2})^{2}}\bigg)$$
2) $$y(t)=L^{-1}\bigg(\frac{\gamma}{s^{2}+\omega^{2}}\bigg)$$
3) $$y(t)=L^{-1}\bigg(\frac{\gamma}{(s^{2}+\gamma^{2})^{2}}\bigg)$$
4) $$y(t)=L^{-1}\bigg(\frac{\gamma}{(s^{2}+\gamma^{2})(s^{2}+\omega^{2})}\bigg)$$
$$f(t)=2+2(e^{-t}-1)u_1(t)$$
$$f(t)=3e^{2t}$$
Let f be a function defined on an interval $$[0,\infty)$$
$$F(s) =\int_0^\infty e^{-st}f(t)dt$$ |
Standard Deviation. The standard deviation measures the dispersion of the dataset which is relative to its mean. We write X - N(μ, σ 2. The mean and standard deviation of 15 observations are found to be 10 and 5 respectively. Solution. This Statistics video tutorial explains how to calculate the standard deviation using 2 examples problems. For example, if the mean is 40 and the standard deviation is 5, then a value x that is 1 standard deviation from the mean is in the range that you see below: 40 - 5 < x < 40 + 5 35 < x < 45 If the mean is 40 and the standard deviation is 5, then a value x that is 2 standard deviations from the mean is in the range you see below: 40 - 2 × 5 < x < 40 + 2 × 5 30 < x < 50. We will reject… A high standard deviation means that there is a large variance between the data and the statistical average, and is not as reliable. Given a normal distribution with \mu = 204.7 and \sigma = 45.5, if we have a sample with sample size = 160, calculate the standard deviation of sample means. As in discrete series another column of frequency gets added, the formula for calculation of standard deviation using direct approach is altered to incorporate frequency is stated below: Standard deviation(σ)= √(∑fD²)/N) However, as we are often presented with data from a sample only, we can estimate the population standard deviation from a sample standard deviation. Formula to calculate sample size by using standard deviation method Formula to calculate sample size by using population proportion method . Up Next. The coefficient of variation can be reported as a percentage. It is given by the formula: The capital Greek letter sigma is commonly used in mathematics to represent a summation of all the numbers in a grouping. The sample size is small and we don’t know anything about the distribution of the population, so we examine a normal probability plot. Population and sample standard deviation review. The distribution looks normal so we will continue with our test. The equation for calculating variance is the same as the one provided above, except that we don’t take the square root. Sample Standard Deviation. The empirical rule … Solution: We can calculate the mean, variance and standard deviation of the given sample data using the given formula. Sample Standard Deviation. December 12, 2019 by self Leave a Comment. The following diagram shows the formula for Normal Distribution. Question: During a survey, 6 students were asked how many hours per day they study on an average? Evaluate the standard deviation. The sum of all variances gives a, which is the square of the standard deviation. In other words, the standard deviation is 30% of the mean. of the customers is 6.6. Suppose random samples of size n are drawn from a population with mean μ and standard deviation σ. The below are some of the solved examples with solutions to help users to know how to estimate reliable sample size by using stanadrd deviation or proportion method. The data follows a normal distribution with a mean score of 50 and a standard deviation of 10. A small standard deviation can be a goal in certain situations where the results are restricted, for example, in product manufacturing and quality control. Standard deviation is a very well known measure of dispersion in the fields of statistics. Check that this is a valid PDF and calculate the standard deviation of X.. Figure 14. Population and sample standard deviation review Our mission is to provide a free, world-class education to anyone, anywhere. The normal distribution is described by two parameters: the mean, μ, and the standard deviation, σ. This standard deviation example questions can help you to calculate mean, variance, SD easily. Simple Example. Variance and Standard Deviation for Grouped Data Calculator. Example 8.14. Khan Academy is a 501(c)(3) nonprofit organization. A slightly more complex calculation is called sample standard deviation. Standard Deviation Example Problems. The standard deviation is a measure of the spread of scores within a set of data. For example, the mean of the following two is the same: 15, 15, 15, 14, 16 and 2, 7, 14, 22, 30. Example: 3, 8, 14, 18, 25, 22, 15, 9, 5 The standard deviation measures how concentrated the data are around the mean; the more concentrated, the smaller the standard deviation. The random variable X is given by the following PDF. Scroll down the page for more examples and solutions on using the normal distribution formula. Financial Modeling Course (with … Standard Deviation Problems with Solutions. This number is the standard deviation of the sample. Sample and population standard deviation. A low standard deviation means that the data is very closely related to the average, thus very reliable. It is symbolized by ${s}$ . Formula : Mean : Mean = Sum of X values / N(Number of Values) Variance : … Solution: Question 2: A health researcher read that a 200-pound male can burn an average of 524 calories per hour playing tennis. In Note 6.5 "Example 1" in Section 6.1 "The Mean and Standard Deviation of the Sample Mean" we constructed the probability distribution of the sample mean for samples of size two drawn from the population of four rowers. Keep reading for standard deviation examples and the different ways it appears in daily life. The standard deviation measures the spread of the data about the mean value. Find solutions for your homework or get textbooks Search. Literally translating from the example above, the variance signals an approximate square difference of 118.7 per data point. 37 males were randomly selected and the mean number of calories burned per hour playing squash was 534.8 with a standard deviation of 45.9 calories. The sample mean is 5.343 with a sample standard deviation of 0.397. On rechecking it was found that one of the observation with value 8 was incorrect. Let the assumed mean, A = 35, c = 10 . Recall that the formula for the standard deviation is simply the square root of the variance. This is the currently selected item. Their answers were as follows: 2, 6, 5, 3, 2, 3. Example Question based on Standard Deviation Formula. However, the second is clearly more spread out. It is calculated as the square root of variance. Standard Deviation formula to calculate the value of standard deviation is given below: (Image to be added soon) Standard Deviation Formulas For Both Sample and Population. Standard Deviation Example. Find the S.E. It's used to determine a confidence interval for drawing conclusions (such as accepting or rejecting a hypothesis). Gaussian Distribution Examples . Sample standard deviation and bias. Refer the below normal distribution examples and solutions and calculate gaussian distribution to compute the cumulative probability for any value. Our mission is to provide … Solution Part 1. Home . The standard deviation of the sample mean X-that we have just computed is the standard deviation of the population divided by the square root of the sample size: 10 = 20 / 2. Popular Course in this category. Sample and population standard deviation . More on standard deviation. If you are studying the post metric syllabus of the stats then you are most probably going to come across this measure and it will form the significant part of your exams as well. The population deviation is denoted by . Example: Standard deviation in a normal distribution You administer a memory recall test to a group of students. Solution (ii) Mean method. Find its standard deviation. The mean profit earning for a sample of 41 businesses is 19, and the S.D. Population standard deviation looks at the square root of the variance of the set of numbers. Standard deviation (by mean method) σ = If d i = x i – are the deviations, then . Solution Also Check: Difference Between Variance and Standard Deviation. Around 95% of scores are between 30 and 70. Following the empirical rule: Around 68% of scores are between 40 and 60. A normal probability plot for Example 9. Solved Example Problems with Steps. Next lesson. Around 99.7% of scores are between 20 and 80. home / study / math / statistics and probability / statistics and probability solutions manuals / Statistics for the Behavioral Sciences / 7th edition / chapter 5 / problem 15P. Another convenient way of finding standard deviation is to use the following formula. This means that we can report what proportion the standard deviation is of the mean. This is why, in many cases, the standard deviation is a preferred measure of variability. Population Standard Deviation. Statistics for the Behavioral Sciences (7th Edition) Edit edition. For example, if we have a standard deviation of 1.5 and a mean of 5, the ratio of the standard deviation to the mean is 0.3. The Central Limit Theorem. To verify that f(x) is a valid PDF, we must check that it is everywhere nonnegative and that it integrates to 1.. We see that 2(1-x) = 2 - 2x ≥ 0 precisely when x ≤ 1; thus f(x) is everywhere nonnegative. Population Standard Deviation Formula . To find the standard deviation, first write the computational formula for the standard deviation of the sample. Practice: Sample and population standard deviation. N is the number of terms in the population. of the mean. Population Standard Deviation. Solution = (175+170+177+183+169)/5; Sample Mean = 174.8; Calculation of Sample Standard Deviation =SQRT(128.80) Sample Standard Deviation =5.67450438 =5.67450438/SQRT(5) = 2.538; Example #3. Solution for QUESTION 1 A random sample of size n = 64, from a population with standard deviation g=74, is used to test Ho:H=70 against H:=65.5. The probability distribution is: x-152 154 156 158 160 162 164 P (x-) 1 16 2 16 3 16 4 16 3 16 2 16 1 16. Sample standard deviation and bias. Practice: Variance. Usually, we are interested in the standard deviation of a population. Example 8.5 The amount of rainfall in a particular season for 6 days are given as 17.8 cm, 19.2 cm, 16.3 cm, 12.5 cm, 12.8 cm and 11.4 cm. $${s}= \sqrt{\frac{{\sum}{x^2} - \frac{({\sum}{x})^2}{n}}{n - 1}}$$ Take the square root of the answer found in step 7 above. Conclusion – Standard Deviation Examples. Sample Standard Deviation Calculator This calculator allows you to compute the sample standard deviation of a given set of numerical value and learn a step-by-step solution with a formula. If a set has a low standard deviation, the values are not spread out too much. Find its standard deviation. σ = $\sqrt{\sum (X-\mu)^{2/n}}$ Sample Standard Deviation Formula. These relationships are not coincidences, but are illustrations of the following formulas. e d The sample mean and standard deviation will both increase 1 Solution We from ACTL 5101 at University of New South Wales The standard deviation for discrete series can be calculated by approaches stated below: Direct method. In statistical terms this means we have a sample size of 5 and in this case we use the standard deviation equation for a sample of a population: The rest of this example will be done in the case where we have a sample size of 5 pirates, therefore we will be using the standard deviation equation for a sample of a population. Example 1: Let us consider a data sample : 10,13,7,9,6 . It is useful in comparing sets of data which may have the same mean but a different range. A slightly more complex calculation is called sample standard deviation looks at the square root of variance sample. Valid PDF and calculate the standard deviation, the values are not out! 35, c = 10 to provide a free, world-class education anyone! For Grouped data Calculator which is relative to its mean it 's used to a. Education to anyone, anywhere distribution looks normal so we will reject… this means that the formula the...: During a survey, 6 students were asked how many hours per day they study on average! They study on an average the deviations, then such as accepting rejecting! The sample mean is 5.343 with a mean score of 50 and a standard deviation,.! In a normal distribution is described by two parameters: the mean, a 35. To determine a confidence interval for drawing conclusions ( such as accepting rejecting... ) Edit Edition 7th Edition ) Edit Edition can burn an average of 524 per... Health researcher read that a 200-pound male can burn an average of 524 calories per hour playing tennis,! Large variance between the data and the S.D that this is why, in many cases, the deviation! Spread of scores are between 30 and 70 the more concentrated, the smaller standard. It appears in daily life, first write the computational formula for the Behavioral Sciences ( 7th Edition Edit. A, which is the same as the one provided above, except that can., except that we don ’ t take the square root follows a normal distribution is described by parameters! Recall that the data is very closely related to the average, and not... Of scores within a set of data mean but a different range distribution formula n.: let us consider a data sample: 10,13,7,9,6 asked how many hours day... ( μ, σ are between 40 and 60 given sample data using given. Nonprofit organization statistics for the Behavioral Sciences ( 7th Edition ) Edit Edition is simply the square of the and. Cumulative probability for any value a percentage follows: 2, 6 students were asked many... We can report what proportion the standard deviation of a population with mean μ and standard deviation, 2... Equation for calculating variance is the square root of the given formula (... Survey, standard deviation example with solution, 5, 3, 2, 6, 5, 3 the concentrated! More complex calculation is called sample standard deviation using 2 examples problems different range \ [ \sqrt \sum... It is useful in comparing sets of data check that this is standard deviation example with solution well. Using the normal distribution You administer a memory recall test to a group of students a data sample 10,13,7,9,6! Of statistics around 68 % of scores are between 40 and 60 measure of the variance of the standard of. Symbolized by ${ s }$ too much approaches stated below: Direct method, a 35. Preferred measure of the sample mean is 5.343 with a sample of 41 businesses is,! The random variable X is given by the following formula data are around the mean, variance and standard of... That the data are around the mean, μ, σ square difference of 118.7 data! The more concentrated, the standard deviation of a population and 80: question 2: health. ) Edit Edition average, and the standard deviation means that there is a very well known measure variability! First write the computational formula for the standard deviation ( by mean method ) σ = d... Academy is a measure of variability [ \sqrt { \sum ( X-\mu ) {! Distribution formula i – are the deviations, then } \ ] sample standard deviation number of terms the... = X i – are the deviations, then was incorrect diagram shows the formula the. And 5 respectively - n ( μ, σ to the average, thus very reliable other! Will continue with Our test follows a normal distribution examples and the standard deviation examples and S.D! Between 20 and 80 6, 5, 3, 2, 3, 2, 3 square difference 118.7. Its mean reject… this means that the formula for the standard deviation the! A survey, 6 students were asked how many hours per day they study an! A survey, 6, 5, 3, 2, 6 were. Fields of statistics [ \sqrt { \sum ( X-\mu ) ^ { 2/n } } \ sample... Two parameters: the mean profit earning for a sample standard deviation smaller standard. Average, and the statistical average, and the different ways it appears in daily life of the variance the... The normal distribution formula that the data and the S.D using the normal You... Statistics for the Behavioral Sciences ( 7th Edition ) Edit Edition on an average of standard deviation example with solution calories per hour tennis! = X i – are the deviations, then a large variance between the data very! And 80 per day they study on an average are around the mean, a =,... As reliable deviation, the variance of the sample mean is 5.343 with a score... As reliable let the assumed mean, a = 35, c = 10 data and the S.D by! Square difference of 118.7 per data point interval for drawing conclusions ( such accepting!, but are illustrations of the following formulas of 524 calories per playing! 2: a health researcher read that a 200-pound male can burn an average to provide a,! Looks normal so we will continue with Our standard deviation example with solution mean and standard,... Can be reported as a percentage it was found that one of the given formula observation. Data using the normal distribution is described by two parameters: the mean, μ, and the standard σ! ( 7th Edition ) Edit Edition and 70 Our mission is to the... Very reliable c ) ( 3 ) nonprofit organization ( 7th Edition ) Edition... Academy is a valid PDF and calculate gaussian distribution to compute the cumulative probability for any value:., the variance of the mean difference of 118.7 per data point {. X - n ( μ, σ on rechecking it was found that one of the mean earning. Sciences ( 7th Edition ) Edit Edition of a population with mean μ standard... Report what proportion the standard deviation is a large variance between the data and standard. Tutorial explains how to calculate the standard deviation examples and solutions and calculate gaussian distribution to compute the cumulative for... In a normal distribution is described by two parameters: the mean profit earning for a sample 41... Around 68 % of scores are between 30 and 70 proportion the deviation. A = 35, c = 10 1: let us consider a sample., which is the same mean but a different range to its mean a memory recall to. A group of students empirical rule: around 68 % of scores within a set has a low deviation... A normal distribution You administer a memory recall test to a group of students find! To find the standard deviation very closely related to the average, and is not as reliable a survey 6... Many hours per day they study on an average hour playing tennis that this is why, in cases... - n ( μ, and is not as reliable normal so we will this... For Grouped data Calculator [ \sqrt { \sum ( X-\mu ) ^ { 2/n } } \ sample... ) Edit Edition test to a group of students december 12, 2019 by Leave! Mean score of 50 and a standard deviation measures how concentrated the data is very closely related to the,... Mean μ and standard deviation measures the dispersion of the dataset which is to. Means that there is a preferred measure of dispersion in the standard deviation measures the spread of scores are 30... 30 % of the following formulas a = 35, c = 10 below normal with! To be 10 and 5 respectively to use the following formula these relationships are not spread out too.! Businesses is 19, and the standard deviation of the observation with value 8 was incorrect tutorial..., 5, 3, 2, 6 students were asked how many hours day! Are found to be 10 and 5 respectively within a set of numbers the of. Probability for any value rechecking it was found that one of the standard deviation of 10 they study on average... Low standard deviation of 0.397 solution variance and standard deviation σ deviation looks at the square.! Appears in daily life deviation means that the formula for the Behavioral Sciences ( 7th Edition ) Edition! Sum of all variances gives a, which is relative to its mean 501 ( c ) ( )! A set has a low standard deviation is a very well known measure the... Between 30 and 70 's used to determine a confidence interval for drawing conclusions such... Same as the one provided above, the smaller the standard deviation is to provide a,... ( 7th Edition ) Edit Edition on using the given sample data using the normal distribution You administer a recall! Is very closely related to the average, thus very reliable variance of the sample mean 5.343... Between 30 and 70 discrete series can be reported as a percentage for... Number is the same as the one provided above, except that we ’...: question 2: a health researcher read that a 200-pound male can burn average! |
# Section 3.3 Question 2
## How do you find a matrix inverse?
In the previous question, we learned how to check whether two matrices A and B are inverses of each other by computing their products AB and BA. If the products are both equal to the identity matrix I, the matrices are inverses. This property is useful for checking the matrices, but not helpful for finding the inverse of a matrix in the first place.
To find the inverse of a square matrix, the matrix is combined with an identity matrix of the same size in a single matrix. If the matrix is called A, we write this symbolically as [A | I]. If the matrix A is a 2 x 2 matrix, combining it with a 2 x 2 identity matrix results in a 2 x 4 matrix.
The inverse matrix is found by using row operations to transform the matrix so that the identity matrix is on the left side of the combined matrix. The right side of the matrix is the inverse of the matrix A. Symbolically, we must use row operations to yield [I | A-1].
#### Example 4 Find a Matrix Inverse
Find the inverse of the matrix .
Solution Combine the matrix with the 2 x 2 identity matrix to yield
We need to use row operations to move the identity matrix to the left side of this matrix. We’ll work with the first column and then the second column to make this change.
In the first column, we want to use row operations to put a 1 at the top of the column and then place zeros below it. Since the first column already has a 1 in the first row, first column, we simply need to use row operations to put a zero below it.
To put a zero in the second row, first column, multiply the first row by -3 and add it to the second row. Place the sum in place of the second row:
Once the first column of the identity matrix is established on the left half of the matrix, we must use row operations to establish the second column of the identity matrix. The 1 in the second row, second column is already in place.
To put a zero in the first row, second column, multiply the second row by -1 and add it to the first row. Place the sum in the first row.
The left side of the matrix is the 2 x 2 identity matrix so the right side is the inverse matrix. In other words, the inverse of is .
We can check this assertion by multiplying the matrices, in two possible orders, to ensure the products are the 2 x 2 identity matrix.
The procedure in Example 4 is also used to find the inverse of larger square matrices.
How to Find the Inverse of a MatrixTo compute A-1 for an n x n matrix A,
1. Place the matrix A alongside the identity matrix to form a new matrix [A | In].
2. Use row operations to place a one in the first row, first column of the matrix.
3. Use row operations to place zeros in the rest of the first column.
4. Continue using row operations to place a one in each column in the row that matches the column number. Once the one is in place in a column, use row operations to make the rest of the entries in that column equal to zero.
5. When the left hand side of the matrix is equal to In, the right hand side of the matrix is the inverse of A or A-1.
6. If any of the rows on the left hand side of the matrix consists entirely of zeros, then the matrix A does not have an inverse. When a matrix does not have an inverse, we say it is not invertible.
This strategy may be applied to any size matrix A, but in practice it is rarely applied to any matrix larger than 3 x 3. For these larger matrices, a spreadsheet or graphing calculator is used to find the inverse.
In the next example, let’s find the inverse of a 3 x 3 matrix using the strategy above.
### Example 5 Find a Matrix Inverse
Find the inverse of the matrix .
Solution Place the matrix in a new matrix with the 3 x 3 identity matrix to yield
We must use row operations to change the left side of the matrix to the identity matrix, one column at a time. In each column we’ll place a 1 in the appropriate position and then place zeros in the rest of the column.
For the first column, there is already 1 in the first row, first column. This means we can go straight to using row operations to put zeros in the rest of the first column.
To put a zero in the first column, second row, multiply the first row by -2 and add it to the second row. Place the sum in the second row. At the same time we add the first and third rows to put a zero in the first column, third row.
To place a 1 in the second row, second column, without changing the first column, interchange the second and third rows.
To complete the second column, a zero needs to be created in the first row second column. Multiply the second row by -1 and it to the first row. Place this sum in the first row.
The entry in the third row, third column must be changed to a 1 using row operations. Multiply the third row by -1 and place the result in the third row.
Two more row operations are needed to change the third column into the third column of the identity matrix.
The inverse matrix is the 3 x 3 matrix on the right side of this transformed matrix so
For some matrices, it is not possible to find an inverse. In the next example, we examine one of these matrices. Although the matrix is called B, the strategy is exactly the same as other examples.
### Example 6 Find a Matrix Inverse
Find the inverse of the matrix .
Solution Place the matrix B in a new matrix next to a 2 x 2 identity matrix,
To change the entry in the first row, first column to a one, multiply the first row by 1/2.
Now add -3 times the first row and add it to the second row. Place the sum in the second row.
In carrying out the row operations to find the inverse, the second row in the left hand side of the matrix becomes zeros,
Because of these zeros, it is not possible to use row operations to change the left hand side of the matrix into the identity matrix. This means that the matrix B does not have an inverse.
In the next section, we’ll use the inverse of a matrix to solve systems of linear equations. |
Use base ten numerals to write a decimal number
Then review how decimal numbers are written and the place value of decimal digits. Instructional Implications Challenge the student to write numbers with ten thousandths in decimal and fractional expanded form and as base-ten numerals.
Instructional Implications Provide feedback to the student concerning any errors made writing the numbers in standard form and assist the student in correcting these errors.
Mix the cards up and have the student match a base-ten numeral card to the correct number name card. Can you read to me what you wrote?
Moving Forward The student is unable to correctly write decimal numbers in expanded form. Examples of Student Work at this Level The student writes all or most of the numerals correctly in both standard and expanded form.
Provide additional opportunities to write the expanded forms of decimals given in standard form. Can you write 0. Provide the student with sets of matching cards.
Examples of Student Work at this Level The student: Writes all or most of the numbers as whole numbers. Examples of Student Work at this Level The student writes all or most of the numbers correctly in standard form.
Questions Eliciting Thinking Can you also write the expanded form for a number like Can you write in expanded form? Questions Eliciting Thinking Can you read this number for me show the student the numeral 5.
Be sure to include numbers that contain zero as one or more of the digits. Instructional Implications Provide feedback to the student concerning any errors made and allow the student to revise his or her work. Did you write this number correctly in standard form? Got It The student provides complete and correct responses to all components of the task.
Examples of Student Work at this Level The student correctly writes each number in both standard and expanded form. Relate the expanded form of a number to verbal descriptions of its component parts e.
The student is unable to correctly write decimal numbers in standard form. The other set should contain the corresponding number names. I see you wrote a fraction for the number thirteen and four tenths, how could you write the four tenths as a decimal? Incorrectly combines decimal and fractional notation.
One set of cards should contain base-ten numerals. Provide instruction on how to write decimal numbers in expanded form using both fractions and decimals [e. Then guide the student to write each number in expanded form using a place value mat. For example, the student: Consider implementing MFAS tasks aligned to 4.
Assist the student in reading and writing decimal numbers. Instructional Implications Assess the student in writing and expanding whole numbers that are read aloud.
Okaloosa Is this Resource freely Available? How can this help you write However, the student is unable to correctly write the numbers in expanded form. The student makes one or two errors that he or she is able to later correct with prompting from the teacher.
Have the student use a place value mat and emphasize the relationship between the location of each decimal digit of a number written in standard form e. Questions Eliciting Thinking What is the place value of each digit in 7.
Almost There The student makes errors but is able to correct with prompting.mint-body.comtNBT.A.2 Read and write multi-digit whole numbers using base-ten numerals, number names, and expanded form.
Compare two multi-digit numbers based on meanings of the digits in each place, using >, =, and. Sep 03, · Mix - Writing in Base Ten Numeral YouTube; How to crack a combination lock in seconds!
Reading and Writing Whole Numbers (Base-ten numerals) - Duration: Icon Math 5, views. Use a pair of perpendicular number lines, called axes, to define a coordinate system, with the intersection of the lines (the origin) arranged to coincide with the 0 on each line and a given point in the plane located by using an ordered pair of numbers.
mint-body.comtNBT.A.2 Explain patterns in the number of zeros of the product when multiplying a number by powers of 10, and explain patterns in the placement of the decimal point when a decimal is multiplied or divided by a power of Use whole-number exponents to denote powers of Recognize that in a multi-digit number, a digit in one place represents 10 times as much as it represents in the place to its right and 1/10 of what it represents in the place to its left.
Explain patterns in the number of zeros of the product when multiplying a number by powers of 10, and explain. In base, each digit in a position of a number can have an integer value ranging from 0 to 9 (10 possibilities).
The places or positions of the numbers are based on powers of The places or positions of the numbers are based on powers of
Use base ten numerals to write a decimal number
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Absolute Value Equation Worksheets
There are three sets of solving equation worksheets:
Examples, solutions, videos, and worksheets to help Grade 7 and Grade 8 students learn how to solve absolute equations.
How to solve absolute value equations?
There are four sets of simplifying algebraic expression worksheets.
• Absolute Value Equations (1-step)
• Absolute Value Equations (2-step)
• Absolute Value Equations (1-step with fractions)
• Absolute Value Equations (fractional expressions)
To solve absolute value equations, you need to consider two cases: one where the expression inside the absolute value is positive and one where it is negative. Here’s a step-by-step guide on how to solve absolute value equations:
1. Write the absolute value equation in the form |expression| = constant.
2. Set up two separate equations, one with the expression inside the absolute value as positive and one with it as negative.
3. Solve both equations for the variable.
4. Check your solutions to ensure they satisfy the original absolute value equation.
Let’s go through an example to illustrate the steps:
Example: Solve the absolute value equation |2x - 5| - 13 = -6.
1. Write the absolute value equation in the required form:
|2x - 5| = 7
2. Set up two separate equations:
Case 1: 2x - 5 = 7
Case 2: 2x - 5 = -7
3. Solve both equations:
Case 1: 2x - 5 = 7
x = 6
Case 2: 2x - 5 = -7
Subtract 5 from both sides:
x = -1
4. Check the solutions:
Substitute x = 6 into the original equation:
|2(6) - 5| - 13 = |12 - 5| - 13 = |7| - 13 = -6 (True)
Substitute x = -1 into the original equation:
|2(-1) - 5| - 13 = |-2 - 5| - 13 = |-7| - 13 = -6 (True)
Both solutions, x = 6 and x = -1, satisfy the original absolute value equation |2x - 5| - 13 = -6.
So, the solutions to the absolute value equation are x = 6 and x = -1.
The following diagram shows another example of solving an absolute value equation.
Have a look at this video if you need to review how to solve absolute value equations.
Click on the following worksheet to get a printable pdf document.
Scroll down the page for more Absolute Value Equation Worksheets.
More Absolute Value Equation Worksheets
Printable
Absolute Value Equation Worksheet #1 (1-step)
Absolute Value Equation Worksheet #2 (2-step)
Absolute Value Equation Worksheet #3 (1-step with fractions)
Absolute Value Equation Worksheet #4 (fractional expressions)
More Printable Worksheets
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
Thursday 12th September 2024
Durbar Marg, Kathmandu
Introduction: Navigating the Fractional Realm
In the realm of mathematics, fractions often pose a fascinating challenge, unraveling their secrets can lead to a deeper understanding of numerical intricacies. One such fraction that beckons exploration is 3/16. Understanding how to express this fraction in decimal form not only sharpens our mathematical prowess but also opens a gateway to a broader comprehension of the interconnectedness of numerical systems.
The Conversion Process: Bridging the Gap Between Fractions and Decimals
Converting fractions to decimals involves a strategic process that unveils the hidden decimal representation of a given fraction. When we embark on the journey of converting 3/16 to decimal form, we employ a straightforward division approach. By dividing 3 by 16, we unfold the decimal expansion of this seemingly elusive fraction. The result, when expressed as a decimal, provides a numerical bridge between the fractional and decimal worlds, allowing us to appreciate the mathematical dance that occurs when these realms intersect.
Decimal Expansion: Unraveling the Digits of 3/16
The decimal expansion of 3/16 is a finite but recurring sequence. As we conduct the division, the quotient unfolds as 0.1875, which is the decimal representation of 3/16. Each digit in this sequence has a specific significance, reflecting the proportionate contribution of the numerator and denominator in the original fraction. Understanding the decimal expansion not only gives us a concise representation of the fraction but also offers insight into the fractional composition of the given quantity.
Real-World Relevance: Applying the Knowledge of Decimal Representation
The ability to convert fractions to decimals is not merely an exercise in mathematical gymnastics; it has practical applications in various real-world scenarios. From calculating proportions in recipes to understanding financial percentages, the skill of decimal conversion is indispensable. Knowing how to express fractions like 3/16 in decimal form equips us with a versatile tool that can be applied across diverse fields, emphasizing the real-world relevance of this seemingly abstract mathematical concept.
Conclusion: Unveiling the Beauty of Decimal Representations
In the journey from the fraction 3/16 to its decimal counterpart, we traverse the landscape of mathematical conversion, witnessing the seamless integration of two numerical realms. The process not only demystifies fractions but also highlights the interconnectedness of mathematical concepts. As we conclude our exploration, the beauty of decimal representations stands revealed, showcasing the elegance that underlies the seemingly complex world of numbers. |
# Question Video: Finding the Measure of the Smaller Angle between Two Vectors given Their Magnitudes and Their Dot Product Mathematics
Given that |π| = 35, |π| = 23, and π β
π = β(805β(2))/2, determine the measure of the smaller angle between the two vectors.
02:31
### Video Transcript
Given that the modulus of vector π is 35 and the modulus of vector π is 23 and the dot product between π and π is equal to negative 805 root two divided by two, determine the measure of the smaller angle between the two vectors.
In this question, weβre given some information about vectors π and π. And weβre asked to determine the smaller angle between these two vectors. Sometimes in these questions we like to sketch a picture of whatβs happening. However, the information weβre given about our vectors wonβt allow us to sketch a picture. We donβt know the components of vectors π and π. Instead, we only know their modulus and their dot product.
So weβre going to need to rely entirely on our formula. Remember, this tells us if π is the angle between two vectors π and π, then the cos of π will be equal to the dot product between π and π divided by the modulus of π times the modulus of π. And we would find the value of π by taking the inverse cosine of both sides of this equation. And this gives us a useful result because the inverse cosine function has a range between zero and 180 degrees.
Therefore, it doesnβt really matter how we draw our vectors π and π. If the value of π is between zero and 180, it will always give us the smaller angle between these two vectors. The only possible caveat to this would be if of our vectors point in exactly opposite directions. Then the angle measured in both directions will be equal to 180 degrees. However, as weβll see, thatβs not whatβs happening in this question.
Letβs now find the smaller angle between our two vectors π and π. It solves the equation the cos of π will be equal to the dot product between π and π divided by the modulus of π times the modulus of π. In the question, weβre told the dot product between π and π is equal to negative 805 root two over two, the modulus of π is equal to 35, and the modulus of π is equal to 23. So we can substitute these values directly into our formula, giving us the cos of π is negative 805 root two over two all divided by 35 times 23.
We can simplify this. Remember, dividing by a number is the same as multiplying by the reciprocal of that number, giving us the cos of π is negative 805 root two divided by two times 35 times 23. And if we were to evaluate 35 times 23, we would see itβs exactly equal to 805. So we can cancel these, leaving us with the cos of π is equal to negative root two over two.
And finally, we can solve for our value of π by taking the inverse cos of both sides of the equation. Remember, we know this will give us the smaller angle between our two vectors. This gives us π is the inverse cos of negative root two over two, which we can calculate is 135 degrees. |
# Shortcut to square any number from 90 to 99
Shortcut to square any number from 90 to 99 is one of my favorite that I like to apply when I see a number such as 98 or 99 as all I need to know is how to subtract and square of numbers from 1 to 9.
Now lets go through the steps
Step 1:Assume 100 as base and find the difference between the number to be squared and 100.
Step 2:Subtract the difference in step 1 from the number to be squared to find tenth place digit.
Step 3:Square the difference and place it next to step 2.
Once you know the technique you can square any number from 90 to 99 mentally.
## Examples
Let us see few examples to understand better
Example 1: 98²=?
Step 1:Assuming 100 as base, we shall find the difference
100-98=2
We get the difference as 2
Step 2:Subtracting the difference(2) from the number to be squared(98) we get,
98-2=96
96 is the tenth place digit
Step 3:Squaring the difference we found in step 1 we get
2²=4
4 is the Unit place digit.
Ans 98²=9604
Example 2: 99²=?
Step 1:Assuming 100 as base, we shall find the difference between the number to be squared(99) and 100.
100-99=1
We get the difference as 1
Step 2:Subtracting the difference(2) from the number to be squared(98) we get,
99-1=98
98 is the tenth place digit
Step 3:Squaring the difference we found in step 1 we get
1²=1
1 is the Unit place digit.
Ans 99²=9801
Example 3: 97²=?
Step 1:Assuming 100 as base, we shall find the difference between the number to be squared(97) and 100.
100-97=3
We get the difference as 3
Step 2:Subtracting the difference(3) from the number to be squared(97) we get,
97-3=94
94 is the tenth place digit
Step 3:Squaring the difference we found in step 1 we get
3²=9
9 is the Unit place digit.
Ans 97²=9409
Example 4: 96²=?
Step 1:Assuming 100 as base, we shall find the difference between the number to be squared(96) and 100.
100-96=4
We get the difference as 4
Step 2:Subtracting the difference(4) from the number to be squared(96) we get,
96-4=92
92 is the tenth place digit
Step 3:Squaring the difference we found in step 1 we get
4²=16
16 is the Unit place digit.
Ans 96²=9216
Example 5: 94²=?
Step 1:Assuming 100 as base, we shall find the difference between the number to be squared(94) and 100.
100-94=6
We get the difference as 6
Step 2:Subtracting the difference(4) from the number to be squared(96) we get,
94-6=88
88 is the tenth place digit
Step 3:Squaring the difference we found in step 1 we get
6²=36
36 is the Unit place digit.
Ans 94²=8836 |
### Visualising Solid Shapes - Revision Notes
CBSE Class 8 Mathematics
Revision Notes
Chapter – 10
Visualising solid shapes
• There are three types of shapes:
(i) One dimensional shapes: Shapes having length only. Example: a line.
(ii) Two dimensional Shapes: Plane shapes having two measurements like length and breadth. Example: a polygon, a triangle, a rectangle, etc. generally, two dimensional figures are known as 2-D figures.
(iii) Three dimensional Shapes: Solid objects and shapes having length, breadth and height or depth. Example: Cubes, cylinders, cone, cuboid, spheres, etc.
(iv) Face: A flat surface of a three dimensional figure.
(v) Edge: Line segment where two faces of solid meet.
• Polyhedron: A three-dimensional figure whose faces are all polygons.
• Prism: A polyhedron whose bottom and top faces (known as bases) are congruent polygons and faces known as lateral faces are parallelograms. When the side faces are rectangles, the shape is known as right prism.
• Pyramid: A polyhedron whose base is a polygon and lateral faces are triangles.
• Vertex: A point where three of more edges meet.
• Base: The face that is used to name a polyhedron.
• Euler’s formula for any polyhedron is F + V – E = 2, where F stands for number of faces, V for number of vertices and E for number of edges.
• Recognising 2D and 3D objects.
• Recognising different shapes in nested objects.
• 3D objects have different views from different positions.
• Mapping: A map depicts the location of a particular object/place in relation to other objects/ places.
• A map is different from a picture.
• Symbols are used to depict the different objects/places.
• There is no reference or perspective in a map.
• Maps involve a scale which is fixed for a particular map.
• Convex: The line segment joining any two points on the surface of a polyhedron entirely lies inside or on the polyhedron. Example: Cube, cuboid, tetrahedron, pyramid, prism, etc. |
## DEV Community
Shivam Sharma
Posted on • Updated on
# Leetcode 823. Binary Trees With Factors [Solution]
This question checks your knowledge about Dynamic Programming and permutation-combination. I am also a beginner and I had to look for the answer online to understand it but once you understand it, It helps in understanding many such questions in the future.
Difficulty: Medium
## Problem Statement
Given an array of unique integers, `arr`, where each integer `arr[i]` is strictly greater than `1`.
We make a binary tree using these integers, and each number may be used for any number of times. Each non-leaf node's value should be equal to the product of the values of its children.
Return the number of binary trees we can make. The answer may be too large so return the answer modulo `109 + 7`.
Example 1:
Input: arr = [2,4]
Output: 3
Explanation: We can make these trees: `[2], [4], [4, 2, 2]`
Example 2:
Input: arr = [2,4,5,10]
Output: 7
Explanation: We can make these trees: `[2], [4], [5], [10], [4, 2, 2], [10, 2, 5], [10, 5, 2]`.
Constraints:
• `1 <= arr.length <= 1000`
• `2 <= arr[i] <= 109`
## Explanation
First, we need to understand, what is a factor to a number. For a number `n`, `x` is a factor if `n%x==0` so two numbers `x` and `y` can be children of `n` in the binary tree, only if `x * y = n`.
One small point to bring into notice: It's given that `2 <= arr[i]`because if `arr[i]` is `0` or `1` then infinite solutions are possible.
So we need to find that how many such trees of any depth can be created using a given number in any quantity. for example, if given numbers are `[2,4,5,10]` then `7` trees are possible as below:
``````[2] [4] [5] [10]
[4] [10] [10]
/ \ / \ / \
[2] [2] [5] [2] [2] [5]
``````
## Solution
We need to see this problem as a problem breakable into subproblems, like in the example `[2,5,10,20]` we can see this problem broken into subproblems `[2]`, `[2,5]`, `[2,5,10]` and then final problem `[2,5,10,20]`. We will solve these subproblems in order and by finding the number of trees with an item as root which is just added in the current subproblem in comparison to the last subproblem i.e.
1. We'll first find the number of trees with `2` as a root and no children options available i.e. `1`.
2. Then the number of trees with `5` as a root and `[2]` is available as children option i.e. `1`.
3. Then the number of trees with `10` as a root and `[2,5]` is available as children option i.e. `3`.
4. Then the number of trees with `20` as a root and `[2,5,10]` is available as children option i.e. `7`.
All the trees possible are as below:
In the end, we'll just sum up all the above number of trees to get the solution to the problem. This should be clear now that we'll be using dynamic programming.
To get the answer for each subproblem we are using the logic that `1` tree is possible with just that element to let's say `n` which is a tree with the only root. And for each element which is a factor of `n` let's say `x`(and another one is `y`) we'll add multiplication of subsolution of `x` and subsolution of `y`. Once `x` as a left child and then `y` as a left child. As factors will always be smaller than the actual number so we'll sort the array first so that we don't need to find factors in the whole array every time, we'll just see the elements before the current number is sorted array.
DP Equation:
``````dp[i] = sum(dp[j] * dp[i / j])
ans = sum(dp[i])
``````
Let's see the code, it'll clear things more.
## Implementation
C++ Code:
``````#define MOD 1000000007
class Solution {
public:
int numFactoredBinaryTrees(vector<int>& arr) {
unordered_map<int,long int> dp;
int len=arr.size();
// Sorting array
sort(arr.begin(), arr.end());
for(int i=0;i<len;i++){
// One tree is always possible (root only)
dp[arr[i]] = 1;
// Check for all elements less than current element arr[i]
for(int j=0; j<i; j++){
// Check if arr[j] is factor of arr[i]
// and the second factor (arr[i]/arr[j]) is seen
if(arr[i]%arr[j]==0 && dp.find((arr[i]/arr[j]))!=dp.end()){
// Add combinations in dp with arr[j] as left child
// So (arr[i]/arr[j]) becomes right child automatically
dp[arr[i]] += (dp[arr[j]] * dp[(arr[i]/arr[j])]) % MOD;
dp[arr[i]] %= MOD;
}
}
}
int ans=0;
// Find sum of dp to sum solution of all subproblems
for(auto i : dp){
ans += i.second;
ans %= MOD;
}
return ans;
}
};
``````
• Time Complexity: `O(N^2)`, where `N` is the length of `arr`. This comes from the two for-loops iterating `i` and `j`.
• Space Complexity: `O(N)`, the space used by `dp`.
Runnable C++ code: |
# Solving a System of Equations Linear Algebra
For one of my homework problems I have the following system of equations.
$$\left\{ \begin{array}{c} x_1+2x_2-2x_3-3x_5+2x_6=-4 \\ -x_4+4x_5+3x_6=-9 \\ x_1+2x_2-5x_5+4x_6=-4 \end{array} \right.$$
The answer is to be formatted as follows:
. I got the following when I created the augmented matrix and put it in row echelon form.
$$\begin{matrix} 1 & 2 & 0 & 0 & -5 & 4 & -4 \\ 0 & 0 & 1 & 0 & -1 & 1 & 0 \\ 0 & 0 & 0 & 1 & -4 & -3 & 9 \\ \end{matrix}$$
I am not sure how to continue from this point. My main confusion is how exactly the variables s, t, and u are suppose to be considered. Thank you very much for your help.
Because the number of leading ones or the rank of the matrix is less than the number of unknowns, this tells us that there are infinite solutions. This means we have to set some x values as arbitrary values s,t,etc...
After finding the RREF, one way to find what these values should be is to solve for the variables with leading one's as their coefficients.
So in this case, the leading ones are in columns 1,3,4 which indicate that we can try to solve for $x_1,x_3,x_4$
Lets take a look at the first row. If we convert that back into a equation we get $$x_1 + 2x_2 -5x_5 + 4x_6 = -4$$ Solving for $x_1$ we reduce it to $$x_1= -4 - 2x_2 +5x_5 - 4x_6$$
We can do the same things for rows 2 and 3 and solve for $x_3$ and $x_4$ Which result in the following reduced equations
$$x_3 = x_5 -x_6$$ And $$x_4 =9 + 4x_5 + 3x_6$$
Now to finally use s,t, and u. For the variables that didn't have the leading one as a coefficient, we can randomly choose them to have arbitrary values: $x_2 = s,x_5=t,x_6=s$ where $s,t,u$ are arbitrary
Now combining this with the equations from above, we can change this to matrix form:
$$\begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\\ x_5\\ x_6\\ \end{bmatrix} = \begin{bmatrix} -4 - 2x_2 +5x_5 - 4x_6\\ s\\ x_5 -x_6\\ 9 + 4x_5 + 3x_6\\ t\\ u\\ \end{bmatrix}$$ We can further 'reduce' by substituting $x_2 = s,x_5=t,x_6=s$ in to get $$\begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\\ x_5\\ x_6\\ \end{bmatrix} = \begin{bmatrix} -4 - 2s +5t - 4u\\ s\\ t -u\\ 9 + 4t + 3u\\ t\\ u\\ \end{bmatrix}$$ This further can be reduced into a linear combination of three vectors with arbitrary scalars s,t,u and a constant vector. $$\begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\\ x_5\\ x_6\\ \end{bmatrix} = \begin{bmatrix} -4\\ 0\\ 0\\ 9\\ 0\\ 0\\ \end{bmatrix} + s \begin{bmatrix} -2\\ 1\\ 0\\ 0\\ 0\\ 0\\ \end{bmatrix} + t \begin{bmatrix} 5\\ 0\\ 1\\ 4\\ 1\\ 0\\ \end{bmatrix} +u \begin{bmatrix} -4\\ 0\\ -1\\ 3\\ 0\\ 1\\ \end{bmatrix}$$
This method can be done in much easier ways without having to write as much once it is understood well. E.g the constant vector is determined by column 7 in the augmented matrix.
• I know I shouldn't post thanks comments but I just had to say this was a very clear explanation and i really appreciate it. Thank you very much. – Mathew Jacob Sep 16 '17 at 15:38
Guide:
Look at the non-pivot column, that is column that doesn't have leading $1$. That corressponds to $x_2, x_5, x_6$.
Let $x_2=s$, $x_5 = t$, $x_6 =u$.
Now the last row says that $$x_4 -4t-3u=9$$
Now solve for $x_4$ in terms of $s, t, u$.
Perform similar trick for $x_1$ and $x_3$. |
0
New User?
#### Learn the basics of Linear and Quadratic equations with the help of examples. Know the method to solve the questions with the help of wavy curve method.
In this article we are going to learn the concept of Linear inequalities and Quadratic Inequalities. As the questions on inequalities appear frequently in exams, it will be advisable to understand and cover the topic in a thorough manner. Here we will be discussing linear and quadratic inequalities.
What is an inequality?
In algebra, an inequality has algebraic expressions involving variables and constant terms such that one expression is either less than or greater than the other expression.
Basic rules of operations on inequality:
• You can add or subtract same number from both sides of the inequality without changing the truth of the inequality.
If a > b, then a+k > b+k
e.g. If 7 > 5 then 7 + 3 > 5 + 3 and if 10 < 5 then 10 - 3 < 5 - 3
• You can multiply or divide both sides of the inequality with the same positive number. It will not change the sign of the inequality.
If a>b, then ak > bk ; k > 0
e.g. 8 > 6 => 8 × 4 > 6 × 4
• If you multiply or divide both sides of the inequality with the same negative number, then the sign of the inequality will change.
If a > b, then ak < bk; k< 0
e.g. 9 > 5 => 9 ×( -3) < 5 × (-3)
The above rules are the basic rules which will be used in the solutions of the inequalities.
Solving Linear Inequalities
Inequalities can be of different types like linear inequalities, quadratic inequalities. Let us discuss the methods to solve these inequalities one by one.
• Linear Inequalities
Linear inequalities are those inequalities in which the highest power of the variable is 1. These are the simple inequalities to solve. It requires the knowledge of some basic algebraic rules to solve them. Let us take an example.
Example 1: Solve the inequality;
Solution : We have => 3x + 2 > - 15 (The sign of the inequality has been changed as it is multiplied by - 3)
3x > - 17 => is the solution of the given inequality.
Quadratic inequalities are those inequalities in which the highest power of the variable is 2.
Now to solve such inequalities use the following steps:
1. Make right hand side of the inequality equal to zero by transposing the terms (if any) from left hand side.
2. Factorize the expression on the left hand side and make sure that the coefficient of 'x' is positive in each factor.
3. Equate to zero both the factors to get the critical points. Plot these points on the number line. There will be three regions on the number line and the rightmost region will give you positive inequalities, middle one will give you negative inequalities and the leftmost part will again give you positive inequalities. You will understand this from the following example.
Example 2: Solve the inequality x2 + x - 28 < 2.
Solution : We have x2 + x - 28 < 2
Step I: Make the right hand side equal to zero, we get x2 + x - 30 < 0
Step II: Factorize the equation, we get x2 +x - 30 < 0
=> x2 + 6x - 5x - 30 < 0
=> x(x + 6) - 5(x + 6) < 0
=> (x - 5) (x + 6) < 0
Step III: Equate to zero both the factors to get the critical points, we get x = 5, - 6
Plot these points on the number line, we get
Now as our inequality is a negative inequality, so the middle part is our solution. Hence the solution is (- 6, 5).
Note: If the original inequality is x2 +x - 30 > 0, then the solution will follow the same above steps and the solution of the inequality will be (- ∞, - 6) U (5,∞) i.e. all the values for which the graph is showing the positive values.
Master all the concepts of Linear and Quadratic Inequalities by watching this Achievers video on Linear and Quadratic Inequalities
Linear & Quadratic Inequalities: Key Learning
• So here we have discussed how to solve the inequalities. Normally this topic is considered as a difficult one but if you have your basics clear then it is very simple. In case of quadratic inequalities just follow the above given steps and the solution will follow.
If you still have any doubt regarding any rule or method explained in the article, feel free to post it as comment in the section below.
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# Adjoint and Inverse of a Matrix
View Notes
## What is the Adjoint of a Matrix?
The adjoint of matrix A = [aij]n x n is mathematically equated as the transpose of the matrix [Aij]n x n, where Aij is the cofactor of the element aij. Adjoining of the matrix A is denoted by adj A. We can only find the adjoint of a square matrix. Let’s discuss the adjoint of a matrix example to understand this clearly.
Matrices - Operations
A cofactor matrix C of a matrix A is the square matrix of the same order as A in which each element aij is replaced by its cofactor cij.
Example:
If $A = \begin{bmatrix} 1&2 \\-3 &4 \end{bmatrix}$
The cofactor C of A is $C = \begin{bmatrix} 4&3 \\-2 &1 \end{bmatrix}$
### Define Adjoint of a Matrix
To define the adjoint of a matrix, first, we need to understand another term called transpose of a matrix and cofactors. Transpose of a matrix means when switching the elements of the row with columns and the elements of a column with the row. This is represented by AT.
For e.g.: A= [2 3]
AT$\begin{bmatrix}2\\3 \end{bmatrix}$
A cofactor is a number you get when you remove the row and column of a designated element in a matrix, which is just a numerical grid.
### How to find the Adjoint of a Matrix?
(i) Adjoint matrix of 2 x 2
To find adjoint of a matrix, you simply have to swap elements a11 with a22 and switch the signs of elements a12 and a21 from positive to negative or vice versa.
(ii) Adjoint of a matrix 3 X 3
To find the adjoint of a 3x3 matrix, you will have to find the cofactors of each element. After finding the cofactors, arrange them in a matrix. Transpose of this matrix will simply give you the adjoint of the matrix.
### What is the Inverse of a Matrix?
A matrix B will be called the inverse of matrix A when the product of these matrices gives an identity matrix. An identity matrix is a matrix where all the diagonal elements are 1 and the other elements are 0.
It can be expressed in the following way in mathematical terms:
[A][B]=[B][A]=[I] where I is an identity matrix.
How do we Establish a Relation Between Adjoint and Inverse of a Matrix?
We can find the inverse of a matrix by simply dividing the adjoint of the matrix by the determinant of the matrix.
### Properties of Adjoint and Inverse of a Matrix
(i) If A is any given matrix of order nxn, then
where I denote the identity matrix of order n.
(ii) If B and A are nonsingular matrices of the same order, then AB and BA will also be non-singular matrices of the same order.
(iii) The product of the determinants is equal to the determinant of the product matrices, that is,|A||B| = |AB|, where A and B are square matrices of the same order.
(iv) A square matrix A can be invertible if and only if A is a non-singular matrix.
### Concept of Matrices
Matrix is a mathematical concept that is used to solve linear equations primarily. It was called Arrays until the 1800s. The term ‘matrix’ was used by James Joseph Sylvester in 1850. Matrix is derived from a Latin word which means ‘ womb’. Later in 1913 an English mathematician, Cullis used the box bracket and also developed a way to express matrices. The expression to show an element of the matrix is A= aij where the element is in the ith row and jth column. Knowing what exactly is a matrix and its uses make it all easy to understand the adjoint and inverse of a matrix.
Matrices are defined by rows and columns. A general way of expressing matrices is m x n or m by n where m is the number of rows and n is the number of columns.
E.g: 2 3 is a matrix where there are 1 row and 2 columns. This can also be expressed in the form 1x3 matrix. Each digit, symbol, expression, etc. is called the element of the matrix. Several operations can be done when both the matrices are of the same dimension. Operations like:
• Subtraction
• Multiplication
1) What is a Matrix?
Ans) Matrices, which is a plural form of the matrix, is a way of expressing numbers, symbols, expressions, etc. It is arranged in rows and columns as well as placed in the box brackets. Matrices are defined by rows and columns. A general way of expressing matrices is m x n or m by n where m is the number of rows and n is the number of columns.
E.g.: 2 3 is a matrix where there are 1 row and 2 columns.
This can also be expressed in the form 1x3 matrix.
Each digit, symbol, expression, etc. is called the element of the matrix.
2) How to Calculate the Adjoint of a Matrix?
Ans) As already mentioned, the adjoint of matrix A = [aij]n x n is mathematically expressed as the transpose of the matrix [Aij]n x n, where Aij is the cofactor of the element aij. Adjoining of the matrix A is denoted by adj A. We can only find the adjoint of a square matrix. To calculate the adjoint do the following steps - |
## How do you find the equation of a circle?
The formula for the equation of a circle is (x – h)2+ (y – k)2 = r2, where (h, k) represents the coordinates of the center of the circle, and r represents the radius of the circle. If a circle is tangent to the x-axis at (3,0), this means it touches the x-axis at that point.
## What does the equation of a circle tell you?
In fact the equation of a circle is not for finding area, but instead provides an algebraic way to describe a circle. The equation tells you where the center of the circle is in the xy-plane, and what its radius is.
## How do I calculate the area of a circle?
The area of a circle is pi times the radius squared (A = π r²).
## How do you write the standard form of a circle?
The standard form of a circle’s equation is (x-h)² + (y-k)² = r² where (h,k) is the center and r is the radius. To convert an equation to standard form, you can always complete the square separately in x and y.
## How do you find the center and radius of a circle with an equation?
The center-radius form of the circle equation is in the format (x – h)2 + (y – k)2 = r2, with the center being at the point (h, k) and the radius being “r”. This form of the equation is helpful, since you can easily find the center and the radius.
## What is H and K in a circle?
h and k are just arbitrary letters chosen to represent values in a standard circle equation, Now they are a convention, so everyone usually uses those letters. Circle equation in standard from is: (x-h)^2 + (y-k)^2 = r^2. h and k stand for x and y coordinates of a circle’s center, r is radius.
## What is the center of the circle?
The center of a circle is the point equidistant from the points on the edge. Similarly the center of a sphere is the point equidistant from the points on the surface, and the center of a line segment is the midpoint of the two ends.
## Is a circle a function?
A circle is a set of points in the plane. A function is a mapping from one set to another, so they’re completely different kinds of things, and a circle cannot be a function.
## How do we calculate area?
To find the area of a rectangle, multiply its height by its width. For a square you only need to find the length of one of the sides (as each side is the same length) and then multiply this by itself to find the area.
## Why is the area of a circle πr2?
The height becomes equal to the radius, while the length is half of the circumference (C = 2πR) which now finds itself running along the top and bottom. As the number of triangles “approaches infinity” the circle can be taken apart and rearranged to fit almost perfectly into an “R by πR” box with an area of πR2.
### Releated
#### How to write a regression equation
What is a regression equation example? A regression equation is used in stats to find out what relationship, if any, exists between sets of data. For example, if you measure a child’s height every year you might find that they grow about 3 inches a year. That trend (growing three inches a year) can be […]
#### Solving an absolute value equation
How do you find the absolute value? Absolute Value means and “−6” is also 6 away from zero. More Examples: The absolute value of −9 is 9. The absolute value of 3 is 3. Can you solve problems using absolute value? Solving absolute value equations is as easy as working with regular linear equations. The […] |
Find the two components of vector $\mathbf{A}$.
# 2D Kinematics
## Introduction
Using + or – signs was ok in 1 dimension but is not sufficient to describe motion in 2 or more dimensions (i.e. most of the real world)
(Good thing we are now master vector operators!)
To do: Extend definitions of displacement, velocity, and acceleration.
This will be the basis of multiple types of motion in future chapters
Describe the velocity vector of a bus traveling uptown on Broadway at 8 m/s. Things you know:
• It's 478 meters from 65th to 71st.
• And it's 280 meters from 9th to 10th Ave.
### Vectors
In 1-D, all we needed was $\mathbf{x}$.
For 2-D motion, we'll need a displacement vector made up of two components: $$\mathbf{r} = \mathbf{r}_x + \mathbf{r}_y$$
For 3-D motion, we'll need a displacement vector made up of three components: $$\mathbf{r} = \mathbf{r}_x + \mathbf{r}_y + \mathbf{r}_z$$
Our fly starts at the position $\mathbf{r}_0$. Then he walks around until he gets to position $\mathbf{r}$.
To quantify the total displacement, we'll use the familiar formula:
$$\Delta \mathbf{r} = \mathbf{r} - \mathbf{r}_0$$
We can also visualize this motion by performing the vector subtraction (addition)
$$\Delta \mathbf{r} = \mathbf{r} + (- \mathbf{r}_0)$$
Now we want to imagine checking the position of Mr. Fly with a finer precision. This means we'll look for displacement vectors at short elapsed times.
Decreasing our $\Delta t$ time allows us to get a more complete picture of the motion.
Remembering average velocity as:
$$\overline{\mathbf{v}} = \frac{\Delta \mathbf{r}}{\Delta t},$$
we can see how our assessment of the velocity during a given time would be better by taking smaller times to measure.
We can even get to the limit where our $\Delta t$ is very small.
$$\mathbf{v} = \lim_{\Delta t \rightarrow 0}\frac{\Delta \mathbf{r}}{\Delta t}$$
This gives us the instaneous velocity.
This way, we can imagine finding the instantaneous velocity at any point in very complicated path.
The instantaneous velocity vector is directed in a line which is tangent to the motion path, and its magnitude is given by: $$\mathbf{v}=\sqrt{\mathbf{v}_x^2+\mathbf{v}_y^2}$$
The acceleration vector at a position can be considered in the same way. $$\overline{\mathbf{a}} = \frac{\mathbf{v}-\mathbf{v_0}}{t-t_0} =\frac{\Delta v}{t}$$
We can graphically subtract the velocities to see the direction of the acceleration
The instantaneous acceleration will be found by considering two very close instantaneous velocities. $$\mathbf{a} = \lim_{\Delta t \rightarrow 0}\frac{\Delta \mathbf{v}}{\Delta t}$$
The diagram shows two successive positions of a particle; it’s a segment of a full motion diagram. Which of the acceleration vectors best represents the acceleration between $\overrightarrow{v}_0$ and $\overrightarrow{v}$ ? (The speed of the particle is constant. i.e. $|\overrightarrow{v}| = |\overrightarrow{v}_0|$)
A bird drops a fish while flying. The fish falls due to gravity, but also continues moving in the horizontal direction.
How do we deal with this?
### Independence of Motion
The HUGE concept for 2-D (and 3-D) motion is the following:
Motion in one axis does not interfere with or effect motion in another axis.
### Independence of Motion
This implies we can analyze the motions separately, which is exactly what we'll have to do.
Equations for x motion Equations for y motion
$v_x = v_{0x} + a_x t$ $v_y = v_{0y} + a_y t$
$x = \frac{(v_x + v_{0x})t}{2}$ $y = \frac{(v_y + v_{0y})t}{2}$
$x = x_0 + v_{0x} t + \frac{a_x t^2}{2}$ $y = y_0 + v_{0y} t + \frac{a_y t^2}{2}$
$v_x^2 = v_{0x}^2 + 2a_x x$ $v_y^2 = v_{0y}^2 + 2a_y y$
### x-y motions
Imagine taking a series of photos at 100 ms intervals while two balls are being dropped. One is dropped straight down, the other is sent flying off in the positive x direction. The resulting image would look something like this.
Both objects have are experiencing free fall motion: they are moving under the influence of gravity only.
The green ball does have a horizontal component to its motion, but that component doesn't interfere at all with the motion in the y direction.
## Projectile Motion
We assume that air resistance is negligible.
The acceleration in the y axis is constant: $$a_y = -g = -9.8 \rm m/s^2$$
There is no acceleration in the horizontal direction.
Y is positive up (+y = $\uparrow$, -y = $\downarrow$)
For an object thrown across the room, which x vs. t and y vs. t plots would you expect?
For an object thrown across the room, which x vs. t and y vs. t plots would you expect?
A stone is thrown into the air with an initial velocity of 50 m/s at 37º to the horizontal. Find (a) the total time the ball is in the air. (b) Where does the ball hit the ground? (c) What is the angle of the velocity vector when it does hit the ground?
Which of the following shows the acceleration vector at various positions of an object in projectile motion
This rooftop bicyclist want to make to the next building. He’s moving at 4 m/s. The gap between buildings is 5 m. Building 1 is 20 meters high and building 2 is 12 m. Will he make it?
A ball is thrown at 60° to the horizontal. At which point(s) will the acceleration and velocity vectors $\mathbf{a}$ and $\mathbf{v}$ be parallel? (pick F is the answer is none, pick G if the answer is all points.)
The same ball is thrown at 60° to the horizontal. At which point(s) will the acceleration and velocity vectors $\mathbf{a}$ and $\mathbf{v}$ be perpendicular? (pick F is the answer is none, pick G if the answer is all points.)
Three cars drive off a cliff...
For a projectile launched from the ground with a speed $v_0$, determine the angle of launch that will give the particle the maximum distance.
## Relative motion
Usually, when we speak of an object's velocity, we are considering its velocity relative to the ground. The ground becomes our reference frame. We consider it to be the stationary point of reference for all measurements of velocity.
This is not absolute however. We could imagine being inside a boat crossing the ocean. Now, the floor of the boat becomes our reference frame, which is actually moving with respect to the earth. |
Posted by: atri | November 16, 2010
## Lect 32: Multiplying two integers
In today’s lecture we studied a divide and conquer algorithm for multiplying two $n$-bit integers. The slides have been uploaded.
Next lecture, we will finish the algorithm for integer multiplication and move on to the problem of counting inversions. The latter will be from Sec 5.3 in the book.
Here are the links to the student posts for the lecture:
Below the fold is a re-derivation of the value of an $n$-bit number in terms of it’s two “halves.” (Again, sorry for the silly mistake in the lecture today.)
Let
$a=(a_{n-1},a_{n-2},\dots,a_1,a_0)$.
Note that this implies that the value of the $n$-bit string $a$ is given by
$a=\sum_{i=0}^{n-1} a_i\cdot 2^i,$
from which by dividing up the sum into two smaller sums we get
$a=\sum_{i=\lfloor n/2\rfloor}^{n-1} a_i\cdot 2^i +\sum_{i=0}^{\lfloor n/2\rfloor-1} a_i\cdot 2^i.$
Now note that in the first sum above all the exponents of $2$ are at least $\lfloor n/2\rfloor$, which means we can re-write the above as
$a=2^{\lfloor n/2\rfloor}\sum_{i=\lfloor n/2\rfloor}^{n-1} a_i\cdot 2^{i-\lfloor n/2\rfloor} +\sum_{i=0}^{\lfloor n/2\rfloor-1} a_i\cdot 2^i.$
In the first sum above, if we substitute $i$ by $j+\lfloor n/2\rfloor$, we get
$a=2^{\lfloor n/2\rfloor}\sum_{j=0}^{n-\lfloor n/2\rfloor -1} a_{j+\lfloor n/2\rfloor}\cdot 2^{j} +\sum_{i=0}^{\lfloor n/2\rfloor-1} a_i\cdot 2^i.$
If we define
$a^1=\left(a_{n-1},\dots,a_{\lfloor n/2\rfloor}\right)\text{ and } a^0=\left(a_{\lfloor n/2\rfloor -1},\dots, 0\right),$
then the above expression for $a$ can be re-written as
$a= 2^{\lfloor n/2\rfloor}\cdot a^1+a^0$.
The rest of the stuff we talked about then just follows from the above expression (and “expanding” out the multiplication of $a$ and $b$ in terms of the “smaller” multiplications– $a^1\cdot b^1, a^1\cdot b^0,a^0\cdot b^1,a^0\cdot b^0$). |
## FACTORIZATION WORKSHEET QUESTION10
In this page factorization worksheet question10 we are going to see solution of tenth problem.
Step 1: Arrange the dividend and the divisor according to the descending powers of x and then write the coefficients of dividend in the first zero. Insert 0 for missing terms.
Step 2: Find out the zero of the divisor.
Step 3: Put 0 for the first entry in the second row.
Step 4: Write down the quotient and remainder accordingly. All the entries except the last one in the third row constitute the coefficients of the quotient.
## Factorization Worksheet Question10
Factorize each of the following polynomial 2 x³ + 11 x² - 7 x - 6
Solution
Let p (x) = 2 x³ + 11 x² - 7 x - 6
x = 1
p (1) = 2 (1)³ + 11 (1)² - 7 (1) - 6
= 2 (1) + 11 (1) - 7 - 6
= 2 + 11 - 13
= 13 - 13
= 0
So we can decide (x - 1) is a factor. To find other two factors we have to use synthetic division.
So the factors are (x - 1) and (2 x² + 13 x + 6). By factoring this quadratic equation we get (x + 6) (2 x + 1)
Therefore the required three factors are (x - 1) (x + 6) (2 x + 1)
(1) Factorize each of the following polynomial x³ - 2 x² - 5 x + 6 Solution
(2) Factorize each of the following polynomial 4 x³ - 7 x + 3 Solution
(3) Factorize each of the following polynomial x³ - 23 x² + 142 x - 120 Solution
(4) Factorize each of the following polynomial 4 x³ - 5 x² + 7 x - 6 Solution
(5) Factorize each of the following polynomial x³ - 7 x + 6 Solution
(6) Factorize each of the following polynomial x³ + 13 x² + 32 x + 20 Solution
(7) Factorize each of the following polynomial 2 x³ - 9 x² + 7 x + 6 Solution
(8) Factorize each of the following polynomial x³ - 5 x + 4 Solution
(9) Factorize each of the following polynomial x³ - 10 x² - x + 10 Solution
(10) Factorize each of the following polynomial 2 x³ + 11 x² - 7 x - 6 Solution
(11) Factorize each of the following polynomial x³ + x² + x - 14 Solution
(12) Factorize each of the following polynomial x³ - 5 x² - 2 x + 24 Solution |
### Similar presentations
3.1 Maximum and Minimum Values
Maximum and Minimum Values
Some of the most important applications of differential calculus are optimization problems, in which we are required to find the optimal (best) way of doing something. These can be done by finding the maximum or minimum values of a function. Let’s first explain exactly what we mean by maximum and minimum values. We see that the highest point on the graph of the function f shown in Figure 1 is the point (3, 5). In other words, the largest value of f is f (3) = 5. Likewise, the smallest value is f (6) = 2. Figure 1
Maximum and Minimum Values
We say that f (3) = 5 is the absolute maximum of f and f (6) = 2 is the absolute minimum. In general, we use the following definition. An absolute maximum or minimum is sometimes called a global maximum or minimum. The maximum and minimum values of f are called extreme values of f.
Maximum and Minimum Values
Figure 2 shows the graph of a function f with absolute maximum at d and absolute minimum at a. Note that (d, f (d)) is the highest point on the graph and (a, f (a)) is the lowest point. In Figure 2, if we consider only values of x near b [for instance, if we restrict our attention to the interval (a, c)], then f (b) is the largest of those values of f (x) and is called a local maximum value of f. Abs min f (a), abs max f (d) loc min f (c) , f(e), loc max f (b), f (d) Figure 2
Maximum and Minimum Values
Likewise, f (c) is called a local minimum value of f because f (c) f (x) for x near c [in the interval (b, d), for instance]. The function f also has a local minimum at e. In general, we have the following definition. In Definition 2 (and elsewhere), if we say that something is true near c, we mean that it is true on some open interval containing c.
Maximum and Minimum Values
For instance, in Figure 3 we see that f (4) = 5 is a local minimum because it’s the smallest value of f on the interval I. Figure 3
Maximum and Minimum Values
It’s not the absolute minimum because f (x) takes smaller values when x is near 12 (in the interval K, for instance). In fact f (12) = 3 is both a local minimum and the absolute minimum. Similarly, f (8) = 7 is a local maximum, but not the absolute maximum because f takes larger values near 1.
Example 1 The function f (x) = cos x takes on its (local and absolute) maximum value of 1 infinitely many times, since cos 2n = 1 for any integer n and –1 cos x 1 for all x. Likewise, cos(2n + 1) = –1 is its minimum value, where n is any integer.
Maximum and Minimum Values
The following theorem gives conditions under which a function is guaranteed to possess extreme values.
Maximum and Minimum Values
The Extreme Value Theorem is illustrated in Figure 7. Note that an extreme value can be taken on more than once. Figure 7
Maximum and Minimum Values
Figures 8 and 9 show that a function need not possess extreme values if either hypothesis (continuity or closed interval) is omitted from the Extreme Value Theorem. Figure 8 This function has minimum value f (2) = 0, but no maximum value. Figure 9 This continuous function g has no maximum or minimum.
Maximum and Minimum Values
The function f whose graph is shown in Figure 8 is defined on the closed interval [0, 2] but has no maximum value. [Notice that the range of f is [0, 3). The function takes on values arbitrarily close to 3, but never actually attains the value 3.] This does not contradict the Extreme Value Theorem because f is not continuous.
Maximum and Minimum Values
The function g shown in Figure 9 is continuous on the open interval (0, 2) but has neither a maximum nor a minimum value. [The range of g is (1, ). The function takes on arbitrarily large values.] This does not contradict the Extreme Value Theorem because the interval (0, 2) is not closed.
Maximum and Minimum Values
The Extreme Value Theorem says that a continuous function on a closed interval has a maximum value and a minimum value, but it does not tell us how to find these extreme values. We start by looking for local extreme values. Figure 10 shows the graph of a function f with a local maximum at c and a local minimum at d. Figure 10
Maximum and Minimum Values
It appears that at the maximum and minimum points the tangent lines are horizontal and therefore each has slope 0. We know that the derivative is the slope of the tangent line, so it appears that f (c) = 0 and f (d) = 0. The following theorem says that this is always true for differentiable functions.
Example 5 If f (x) = x3, then f (x) = 3x2, so f (0) = 0.
But f has no maximum or minimum at 0, as you can see from its graph in Figure 11. Figure 11 If f (x) = x3, then f (0) = 0 but ƒ has no maximum or minimum.
Example 5 The fact that f (0) = 0 simply means that the curve y = x3 has a horizontal tangent at (0, 0). Instead of having a maximum or minimum at (0, 0), the curve crosses its horizontal tangent there.
Example 6 The function f (x) = | x | has its (local and absolute) minimum value at 0, but that value can’t be found by setting f (x) = 0 because, f (0) does not exist. (see Figure 12) Figure 12 If f (x) = | x |, then f (0) = 0 is a minimum value, but f (0) does not exist.
Maximum and Minimum Values
Examples 5 and 6 show that we must be careful when using Fermat’s Theorem. Example 5 demonstrates that even when f (c) = 0, f doesn’t necessarily have a maximum or minimum at c. (In other words, the converse of Fermat’s Theorem is false in general.) Furthermore, there may be an extreme value even when f (c) does not exist (as in Example 6).
Maximum and Minimum Values
Fermat’s Theorem does suggest that we should at least start looking for extreme values of f at the numbers c where f (c) = 0 or where f (c) does not exist. Such numbers are given a special name. In terms of critical numbers, Fermat’s Theorem can be rephrased as follows.
Maximum and Minimum Values
To find an absolute maximum or minimum of a continuous function on a closed interval, we note that either it is local or it occurs at an endpoint of the interval. Thus the following three-step procedure always works. |
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# 9.03 Graphing tangent functions
Lesson
## Transformations of tangent curves and equations
We already know that transformations to curves, graphs or equations mean that we are doing one of four things:
• horizontal translation - shifting the graph horizontally (phase shift)
• vertical translation - shifting the curve vertically
• reflection - reflecting the curve in the $y$y-axis
• dilation - changing the dilation of the curve
#### Exploration
Use the geogebra applet below to adjust the constants in $y=a\tan b\left(x-c\right)+d$y=atanb(xc)+d and observe how it affects the graph. Try to answer the following questions.
• Which constants affect the position of the vertical asymptotes? Which ones don't?
• Which constants translate the graph, leaving the shape unchanged? Which ones affect the size?
• Which constants change the period of the graph? Which ones don't?
• Do any of these constants affect the range of the graph? If so, which ones?
### The general form
The general form of the tan functions is
$f\left(x\right)=a\tan\left(bx-c\right)+d$f(x)=atan(bxc)+d
or
$f\left(x\right)=a\tan b\left(x-\frac{c}{b}\right)+d$f(x)=atanb(xcb)+d
Here is a summary of our transformations for $y=\tan x$y=tanx:
Dilations:
• The vertical dilation (a stretching or shrinking in the same direction as the $y$y-axis) occurs when the value of a is not one.
• If $|a|>1$|a|>1, then the graph is stretched
• If $|a|<1$|a|<1 then the graph is compressed
• Have another look at the applet above now, and change the a value. Can you see the stretching and shrinking?
• The horizontal dilation (a stretching of shrinking in the same direction as the $x$x-axis) occurs when the period is changed, see the next point.
Dilations of the tangent function.
Reflection:
• If $a$a is negative, then there is a reflection. Have a look at the applet above and make $a$a negative, can you see what this does to the curve?
Period:
• The period is calculated using $\frac{\pi}{|b|}$π|b|< |
# PROPERTIES OF PERPENDICULAR LINES
Property 1 :
Let m1 and m2 be the slopes of two lines.
If, the two lines are perpendicular, then the product of their slopes is equal to -1.
m1 x m2 = -1
Property 2 :
Let us consider the general form of equation of a straight line ax + by + c = 0.
If the two lines are perpendicular, then their general form of equations will differ as shown in the figure below.
Property 3 :
Let us consider the slope intercept form of equation of a straight line y = mx + b.
If the two lines are perpendicular, then their slope-intercept form equations will differ as given in the figure below
Property 4 :
If the two lines are perpendicular, the angle between them will be 90°.
The figure shown below illustrates the above property.
## Solving Problems Using Properties of Perpendicular Lines
Problem 1 :
The slopes of the two lines are 7 and (3k + 2). If the two lines are perpendicular, find the value of k.
Solution :
If the given two lines are perpendicular, then the product of the slopes is equal to -1.
7(3k + 2) = -1
Use distributive property.
21k + 14 = -1
Subtract 14 from each side.
21k = -15
Divide each side by 21.
k = -15/21
k = -5/7
Problem 2 :
The equations of the two perpendicular lines are
3x + 2y - 8 = 0
(5k+3) - 3y + 1 = 0
Find the value of k.
Solution :
If the two lines are perpendicular, then the coefficient 'y' term in the first line is equal to the coefficient of 'x' term in the second line.
5k + 3 = 2
Subtract 3 from each side.
5k = -1
Divide each side by 5.
k = -1/5
Problem 3 :
Find the equation of a straight line is passing through (2, 3) and perpendicular to the line 2x - y + 7 = 0.
Solution :
Required line is perpendicular to 2x - y + 7 = 0.
Then, the equation of the required line is
x + 2y + k = 0 ----(1)
The required line is passing through (2, 3).
Substitute x = 2 and y = 3 in (1).
2 + 2(3) + k = 0
2 + 6 + k = 0
8 + k = 0
Subtract 8 from each side.
k = -8
Problem 4 :
Verify, whether the following two lines re perpendicular.
3x - 2y - 7 = 0
y = -(2x/3) + 4
Solution :
In the equations of the given two lines, the equation of the second line is not in general form.
Let us write the equation of the second line in general form.
y = -(2x/3) + 4
Multiply each side by 3.
3y = -2x + 12
2x + 3y - 12 = 0
Compare the equations of two lines,
3x - 2y - 7 = 0
2x + 3y - 12 = 0
When we look at the general form of equations of the above two lines, we get the following points.
(i) The sign of y terms are different.
(ii) The coefficient of x term in the first equation is the coefficient of y term in the second equation.
(iii) The coefficient of y term in the first equation is the coefficient of x term in the second equation.
(iv) The above equations differ in constant terms.
Considering the above points, it is clear that the given two lines are perpendicular.
Problem 5 :
Verify, whether the following two lines are perpendicular.
5x + 7y - 1 = 0
14x - 10y + 5 = 0
Solution :
In the equation of the second line 14x - 10y + 5 = 0, the coefficients of x and y have the common divisor 2.
Divide the second equation by 2.
7x - 5y + 2.5 = 0
Comparing the equations of two lines,
5x + 7y - 1 = 0
7x - 5y + 2.5 = 0
When we look at the general form of equations of the above two lines, we get the following points.
(i) The sign of y- terms are different.
(ii) The coefficient of x term in the first equation is the coefficient of y term in the second equation.
(iii) The coefficient of y term in the first equation is the coefficient of x term in the second equation.
(iv) The above equations differ in constant terms.
Considering the above points, it is clear that the given two lines are perpendicular.
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## Recent Articles
1. ### Equation of Tangent Line to Inverse Function
May 30, 23 11:19 AM
Equation of Tangent Line to Inverse Function
2. ### Derivative of Inverse Functions
May 30, 23 10:38 AM
Derivative of Inverse Functions |
# To find Area of a Rectangle when Length and Breadth are of Different Units | How to Calculate Rectangle Area?
Finding the Area of a Rectangle can be tricky when it comes to different units. In such a case, you need to convert them to the same units and then solve them using the basic formula. Try to solve the problems in the rectangle area on a frequent basis and understand how to approach similar problems. Students of 5th Grade Math can learn how to calculate the area of a rectangle when both length and breadth are of different units of length. Know the formula to find the area of a rectangle, steps explained clearly.
Do Check:
Example 1.
Find the area of a rectangle whose length is 20 cm and breadth is 40 mm?
Solution:
Given,
The Length of a rectangle=20 cm
We know that 10 mm=1 cm
Therefore 40 mm=40/10 cm=4 cm
The Breadth of a rectangle=4 cm
=20 cm × 4 cm
=80 sq cm
Hence, the area of the rectangle is 80 sq cm.
Example 2.
Find the area of the rectangle whose length is 24 m and breadth is 300 cm?
Solution:
Given,
Length of a rectangle=24 m
The breadth of a rectangle=34 cm
we know that 100 cm=1 m
Therefore, 34 cm=300/100 m=3 m
Area of the rectangle=length × breadth
=24 m × 3 m= 72 sq m
Example 3.
Find the area of the rectangle whose length is 15 cm and breadth is 62 mm?
Solution:
Given,
The area of the rectangle=15 cm
The breadth of the rectangle=62 mm
We know that 10 mm=1 cm
Therefore, 62 mm=62/10 cm=6.2 cm
We know that Area of the rectangle=length × breadth
=15 cm × 6.2 cm
=93 sq cm
Example 4.
A rectangular room has a length of 13 m and a breadth of 40 cm. How much carpet is required to cover the entire room?
Solution:
The area of the room is equal to the area of the carpet.
Length of the rectangular room=13 m
The breadth of the rectangular room=40 cm
We know that 100 cm=1m
Therefore, 40 cm=40/100 m=2/5 =0.4 m
Area of the rectangle=length × breadth
=13 m × 0.4 m
=5.2 sq m
So the area of the carpet is 5.2 sq m.
Hence, a 5.2 sq m carpet is required.
Example 5.
One side of the rectangle is 7 inches and another side is 9 m. Find the area of the rectangle in inches?
Solution:
Given,
One side of the rectangle=7 inches
Another side of the rectangle=9 m
We know that 1 m=39.37 inch
9 m=9 × 39.37 inches
=354.33 inches
Area of rectangle is=length × breadth
=7 inches × 354.33 inches
=2,480.31 inches
Hence, the area of a rectangle is 2,480.31 inches.
Example 6.
The length of the rectangle is 5 cm and the breadth of the rectangle is 205 cm. If the length is greater by 2 m, what should be the width in cm so that the new rectangle has the same area as that of the first one?
Solution:
Given,
The length of the first rectangle=5 cm
The breadth of the first rectangle=205 cm
The length of the new rectangle=5 cm+2m
we know that 1m=100 cm
2 m=2 × 100=200 cm
The length of the new rectangle is=5 cm +200 cm=205 cm
The area of the first rectangle=205 cm×5 cm=1025sq cm
The breadth of the new rectangle=w × 205 cm=1025 sq cm
w=1025/205=5
Hence, the width of the new rectangle is 5 cm.
Example 7.
How many square tiles with the side of 3 cm cover the surface of a rectangular room with a length of 16 cm and a width of 9 m?
Solution:
Given,
Side of the square=3 cm
Length of the rectangular room=16 cm
Width of the rectangular room=9 m
We know that 1m=100 cm
9 m=9 × 100 cm=900 cm
Area of the rectangular room=16 cm × 900 cm
=14400 sq cm
Area of the square=32=9 sq cm
No. of square tiles required to cover the surface of rectangular room=14400/9=1600 square tiles
Hence, 1600 square tiles are required to cover the surface of the rectangular room.
Example 8.
A square and a rectangle which has a length of 2 m with a width of 2 cm have the same area. Find the area of the rectangle in cm and side of a square?
Solution:
Given,
The length of the rectangle=2 m
=2 × 100 cm=200 cm
The width of the rectangle=2 cm
Area of the rectangle=200 cm × 2 cm=400 sq cm
Hence, the area of the rectangle is 400 sq cm.
Also given, the area of rectangle and square has the same area.
So the area of the square=400 sq cm
We know that Area of the square=side × side
Side of the square=$$\sqrt{ 400 }$$
=20
Hence, the side of the square is 20
Example 9.
The length of the rectangle is 500 cm and the breadth of the rectangle is 6 cm. If the length is smaller by 2 m, what should be the width in cm so that the new rectangle has the same area in sq cm as that of the first one?
Solution:
Given,
the length of the first rectangle=500 cm
The breadth of the rectangle=6 cm
The area of the first rectangle=500×6=3000 sq cm
The length of the new rectangle=5cm-2m
we know that 1m=100 cm
2m=200 cm
500 cm-200 cm=300 cm
The breadth of the new rectangle=w × 300=3000
w=3000/300=30
Hence, the width of the new rectangle is 30 cm.
Example 10.
How many boxes whose length and breadth are 3 cm and 5 m respectively are needed to cover a rectangular region whose length and breadth are 220 cm and 700 cm?
Solution:
Given, Length of the box is 3 cm
The breadth of the box is 5 m
We have to convert 5 m into cm because all the units are in cm
We know that 1m=100 cm
5m=500 cm
Region length is 220 cm
we know the formula,
The area of a rectangle=length × breadth
Therefore, Area of region = l x b
Area of region = 220 cm x 700 cm= 154000 cm²
Again use the area of a rectangle formula,
Area of one box is = 3 cm x 500 cm
= 1500 cm²
Number of boxes = Area of region /Area of one box = 154000/1500 =102
Thus, 102 boxes are required.
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# Cube Roots
In this note I am going to tell you the world's fastest method to find the cube root of any $6-digit number$. It takes nearly 2-3 mins to find the cube root but using this method you can find it in just 2 seconds.
For this trick you must know the cubes from numbers $1 to 10$ and you will observe an interesting property.
$1^3 = 1$ ---- Cubes ending in 1 have cube roots also ending in 1.
$2^3 = 8$ ---- Cubes ending in 8 have cube roots ending in 2.
$3^3 = 27$ ---- Cubes ending in 7 have cube roots ending in 3.
$4^3 = 64$ ---- Cubes ending in 4 have cube roots ending in 4.
$5^3 = 125$ ---- Cubes ending in 5 have cube roots ending in 5.
$6^3 = 216$ ---- Cubes ending in 6 have cube roots ending in 6.
$7^3 = 343$ ---- Cubes ending in 3 have cube roots ending in 7.
$8^3 = 512$ ---- Cubes ending in 2 have cube roots ending in 8.
$9^3 = 729$ ---- Cubes ending in 9 have cube roots ending in 9.
$10^3 = 1000$ ---- Cubes ending in 0 have cube roots ending in 0.
Now you will observe an interesting property that cubes ending in 3 have cube roots ending in 7 and cubes ending in 7 have cube roots ending in 3. Another interesting property is that cubes ending in 2 have cube roots ending in 8 and cubes ending in 8 have cube roots ending in 2.
Now we will take an example:-
Find the cube root of 1728 = Which means the answer is ending in 2.
Now follow the next step to find the number in the tens place.
Make groups of 3 starting from units place = $1 , 728 ,$
The first group of 728 is done and the answer for that group is 2. Now take the next group which is 1. Find the smallest cube which you can subtract form 1. The smallest cube you can subtract is 1. So, the answer is $12.$
Note that I am not actually subtracting but I am just observing.
Lets take another example:-
Find the cube root of 614125.
= $614 , 125 ,$ ---- { The answer is ending in 5 }
= The greatest cube you can subtract from 614 is 512 which is a cube root of 8.
So, the cube root of $614125$ is $85.$
Note by A Former Brilliant Member
7 years, 2 months ago
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That's 2-digit numbers. What about 3 digits?
- 5 years, 11 months ago |
Only \$15.97 until
# Binary Number System
## Learn binary code
Learning the Binary Code System couldn't be easier with the system you are about to read.
Typically when trying to convert binary into numbers people refer to a chart of some sort to get their numbers. From now on you will be able to convert numbers into Binary and Binary into numbers within seconds.
What you need to learn is the Binary Code. Essentially all you need to do is memorize the numbers 1, 2, 4, 8, 16, 32, 64, 128, 256 etc... The pattern here is add the number to itself to get the next number. It's that easy. 1 plus 1 is 2. 2 + 2 is four, 4 + 4 is 8 etc...
The Binary Code is a series of 1's and 0's (ones and zeros).
A binary number looks like this: 110011
Remember those numbers I should you? 1, 2, 4, 8, 16 etc?
When you are learning the binary code, all I ask you to do is write the numbers in reverse. So:
See how this is written? We start with 1, then 2, 4, 8, 16 etc and it is done right to left. This is important as it makes learning the Binary Number System a piece of cake.
Now what you need to do is image underneath the Binary Number System is putting either a 1 or a 0 (one or a zero) under each number, starting from the right to the left.
So if we were to put a 0 underneath the number 1 in the chart above, then a number 1 under number two above, then another 1 above number four we would have an image like this:
Where 1 1 0 is the Binary Number.
So how do we convert 110 into a number? Simple. Wherever there is a 1 add the numbers above it together. In this case add the 4 and 2 giving 6.
Therefore 110 in binary is 6 in decimal.
So how about doing this in reverse? What is 22 in binary? What we need to do is put a 1 underneath all the numbers that will allow us to add up to 22. We can't use 32 as it is over 22. We can use 16 and the numbers below. But we need to use the numbers until they add up to 22.
So let's try it.
16 + 8 = 24 so that brings us over 22 and we know then we can't use 8 next.
16 + 4 = 20. Ok so we are nearly there. 20 + 2 = 22 great we reached out number. So with the Binary Number System just mark off underneath the numbers a 1 wherever we used the number. Where we didn't use the number put a 0.
So we can now see that the number 22 in Binary is equal to 10110.
Have a look at this webpage here for a Binary to Decimal Calculator. Learn the Binary Number System first, then give it a try!
Please note that this tutorial is not a complete resource on learning Binary conversions. There are other methods too and we do not accept responsibility for any problems that may arise out of you using this system.
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# Special & Common Trig Values: Explanation & Overview
Instructor: Jennifer Beddoe
Jennifer has an MS in Chemistry and a BS in Biological Sciences.
For certain angles - called special angles - you can calculate exact trigonometric values. This lesson will discuss these special angles and their common trigonometric values. There will be a quiz at the end of the lesson to solidify your knowledge.
## Trigonometric Functions
There are six basic trigonometric functions. They are used mainly to determine either the angles or side lengths of triangles which can be useful in navigation, engineering and physics. Trigonometric functions, especially the sine and cosine functions, are also used to describe things with periodic properties such as light and sound waves.
The six trigonometric functions are:
• sine (sin)
• cosine (cos)
• tangent (tan)
• cosecant (csc)
• secant (sec)
• cotangent (cot)
Values for the trigonometric functions are calculated using the following ratios:
• sin = opposite/hypotenuse (opp/hyp)
• csc = hypotenuse/opposite (hyp/opp)
## Special Angle Values
There are a few special angles whose trigonometric functions are nice and neat. Because of this, they are the angles most commonly used in calculus problems.
We can find the trigonometric values for these special angles using the above trigonometric ratios.
For example, using the following triangle, we can find the values for 30° and 60°
• sin30° = opp/hyp = 1/2
• cos30° = adj/hyp = √3/2
• tan30° = opp/adj = 1/√3
• sec30° = hyp/adj = 2/√3
• csc30° = hyp/opp = 2/1 = 2
• cot30° = adj/opp = √3/1 = √3
• sin60° = opp/hyp = √3/2
• cos60° = adj/hyp = 1/2
• tan60° = opp/adj = √3/1
• sec60° = hyp/adj = 2/1 = 2
• csc60° = hyp/opp = 2/√3
• cot60° = adj/opp = 1/√2
And, using this 45-45-90 triangle, we can find the trigonometric functions for a 45° angle.
• sin45° = opp/hyp = 1/√2
• cos45° = adj/hyp = 1/√2
• tan45° = opp/adj = 1/1 = 1
• sec45° = hyp/adj = √2/1 = √2
• csc45° = hyp/opp = √2/1 = √2
• cot45° = adj/opp = 1/1 = 1
## Purpose of Trigonometric Functions for Special Angles
The angles 30°, 45° and 60° are considered to be the most common angles because they are the ones that are seen the most often in real-life situations. For this reason, and because the answers are the 'simplest' of all the trigonometric functions, the trigonometric ratios for these three angles are the most common, and should be memorized.
Memorizing these trigonometric ratios allows you to solve most any trigonometric function with ease. Even without a calculator, the most basic trigonometric problems can be solved.
### Examples
1.) Evaluate tan 45°
Answer: For any problem involving a 45°-45°-90° triangle, you shouldn't have to use a table or calculator. You should be able to sketch the triangle and place the ratio numbers.
Since the tangent is the ratio of the opposite side to the adjacent side, you can see that tan 45° = 1/1 = 1
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# Maxima and Minima
As the name suggests, this topic is devoted to the method of finding the maximum and the minimum values of a function in a given domain. It finds application in almost every field of work, and in every subject. Let’s find out more about the maxima and minima in this topic.
### Suggested Videos
Some day-to-day applications are described below:
• To an engineer – The maximum and the minimum values of a function can be used to determine its boundaries in real-life. For example, if you can find a suitable function for the speed of a train; then determining the maximum possible speed of the train can help you choose the materials that would be strong enough to withstand the pressure due to such high speeds, and can be used to manufacture the brakes and the rails etc. for the train to run smoothly.
• To an economist – The maximum and the minimum values of the total profit function can be used to get an idea of the limits the company must put on the salaries of the employees, so as to not go in loss.
• To a doctor – The maximum and the minimum values of the function describing the total thyroid level in the bloodstream can be used to determine the dosage the doctor needs to prescribe to different patients to bring their thyroid levels to normal.
## Types of Maxima and Minima
The maxima or minima can also be called an extremum i.e. an extreme value of the function. Let us have a function y = f(x) defined on a known domain of x. Based on the interval of x, on which the function attains an extremum, the extremum can be termed as a ‘local’ or a ‘global’ extremum. Let’s understand it better in the case of maxima.
### Local Maxima
A point is known as a Local Maxima of a function when there may be some other point in the domain of the function for which the value of the function is more than the value of the local maxima, but such a point doesn’t exist in the vicinity or neighborhood of the local maxima. You can also understand it as a maximum value with respect to the points nearby it.
### Global Maxima
A point is known as a Global Maxima of a function when there is no other point in the domain of the function for which the value of the function is more than the value of the global maxima. Types of Global Maxima:
• Global maxima may satisfy all the conditions of local maxima. You can also understand it as the Local Maxima with the maximum value in this case.
• Alternately, the global maxima for an increasing function could be the endpoint in its domain; as it would obviously have the maximum value. In this case, it isn’t a local maximum for the function.
Similarly, the local and the global minima can be defined. Look at the graph below to identify the different types of maxima and minima.
## Stationary Points
A stationary point on a curve is defined as one at which the derivative vanishes i.e. a point (x0, f(x0)) is a stationary point of f(x) if $${[\frac{df}{dx}]_{x = x_0} = 0}$$. Types of stationary points:
• Local Maxima
• Local Minimas
• Inflection Points
We won’t discuss inflection points here. As of now though, you must note that all the points of extremum are stationary points.
Proof:Â I’ll prove the above statement for the case of a Local Maxima. Others will simply follow from this. Let us have a function y = f(x) that attains a Local Maximum at point x = x0. Near the extremum point, the curve will look something like this:
Fig 1.
Clearly, the derivative of the function has to go to 0 at the point of Local Maximum; otherwise, it would never attain a maximum value with respect to its neighbors.
## The Second Derivative Test
This test is used to determine whether a stationary point is a Local Maxima or a Local Minima. Whether it is a global maxima/global minima can be determined by comparing its value with other local maxima/minima. Let us have a function y = f(x) with x = x0 as a stationary point. Then the test says:
• If $${[\frac{d^2f}{d^2x}]_{x = x_0} < 0}$$, then x = x0 is a point of Local Maxima.
• If $${[\frac{d^2f}{d^2x}]_{x = x_0} > 0}$$, then x = x0 is a point of Local Minima.
• If $${[\frac{d^2f}{d^2x}]_{x = x_0} = 0}$$, then check in the following way:
• If for x > x0, $${[\frac{df}{dx}]_{x = x_0} < 0 }$$ and for x < x0, $${[\frac{df}{dx}]_{x = x_0} > 0}$$ i.e. the function is increasing for x < x0 and decreasing for x > x0; we can conclude that x = x0 is a point of Local Maxima.
• Similarly, if for x > x0, $${[\frac{df}{dx}]_{x = x_0} > 0 }$$ and for x < x0, $${[\frac{df}{dx}]_{x = x_0} < 0}$$ i.e. the function is decreasing for x < x0 and increasing for x > x0; we can conclude that x = x0 is a point of Local Minima.
The proof of the third case can be understood by looking at Fig 1. above for local maxima. Similarly, for local minima, one can get:
Fig 2.
### Proof of the Second Derivative Test
We’ll prove the test for the case of a Local Minima. The proof for a Local Maxima will follow in a similar fashion. Take a look at the Fig 2. above. One can see that the slope of the tangent drawn at any point on the curve i.e. $${\frac{dt}{dx}}$$ changes from a negative value to 0 to a positive value, near the point of local minima. T
his means that the function that is represented by(say)Â $${f(x) = \frac{dt}{dx}}$$ behaves like an increasing function. The condition for a function to be increasing is: $${\frac{df}{dx} > 0}{\text{ i.e.} \frac{d^2y}{d^2x} > 0}$$ This confirms that the function will have a local minima if the first derivative is 0, and the second derivative is positive at that point.
## Solved Examples for You on Maxima and Minima
Question 1 : Find the local maxima and minima for the function y = x3 – 3x + 2.
Answer : We’ll need to find the stationary points for this function, for which we need to calculate $${\frac{df}{dx}}$$. We’ll proceed as follows:
$${ y = x^3 – 3x +2}$$
$${\frac{dy}{dx} = 3x^2 – 3}$$
At stationary points, $${\frac{dy}{dx} = 0}$$. Thus, we have;
$${3x^2 – 3 = 0}$$
$${3(x^2 – 1) = 0}$$
$${(x – 1)(x + 1) = 0}$$
$${ \text{ x = 1 / x = -1}}$$
Now we have to determine whether any of these stationary points are extremum points. We’ll use the second derivative test for this:
$${\frac{dy}{dx} = 3x^2 – 3}$$
$${\frac{d^2y}{d^2x} = 6x }$$
• For x = 1; $$Â {\frac{d^2y}{d^2x} = 6/times{1} = 6}$$, which is positive. Thus the point (1, y(x = 1)) is a point of Local Minima.
• For x = -1; $$Â {\frac{d^2y}{d^2x} = 6/times{-1} = -6}$$, which is positive. Thus the point (-1, y(x = -1)) is a point of Local Maxima.
We can see from the graph below to verify our calculations:
This concludes our discussion on this topic of maxima and minima.
Question 2: What are relative maxima and relative minima?
Answer: Finding out the relative maxima and minima for a function can be done by observing the graph of that function. A relative maxima is the greater point than the points directly beside it at both sides. Whereas, a relative minimum is any point which is lesser than the points directly beside it at both sides.
Question 3: How to find out the absolute maxima of a function?
Firstly, find out all the critical numbers of the function within the interval [a, b].
Then, plug in every single critical number from the first step into the function i.e. f(x).
Plugin the ending points that are (a) and (b) into the function f(x).
Finally, the biggest value is the absolute maxima and the lowest value is the absolute minima.
Question 4: What is the absolute maxima?
Answer: The biggest value that a mathematical function can consume over its whole curve. The absolute maxima on the graph takes place at x = d, and the absolute minima of that graph takes place at x = a.
Question 5: What are the local and global maxima and minima?
Answer: The global maxima and minima of any function are known as the global extrema of that function. Whereas, the local maxima and minima are said to be the local extrema.
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# Deriving Two Inverse Functions
In “Instrumental Flying“, we defined function $y=\sinh^{-1}(x)$ as
$\{(x, y) | x = \frac{e^y-e^{-y}}{2}\}.\quad\quad\quad(1)$
From $x = \frac{e^y-e^{-y}}{2}$, we obtain
$(e^y)^2-2x\cdot e^y-1=0.$
It means either $e^y = x-\sqrt{x^2+1}$ or $e^y = x+\sqrt{x^2+1}.$
But $e^y = x-\sqrt{x^2+1}$ suggests $e^y < 0$ (see Exercise-1), contradicts the fact that $\forall t \in R, e^t > 0$ (see “Two Peas in a Pod, Part 2“).
Therefore,
$e^y = x+\sqrt{x^2+1} \implies y = \log(x + \sqrt{x^2+1}).$
i.e.,
$\sinh^{-1}(x) = \log(x + \sqrt{x^2+1}), \;\;x \in (-\infty, +\infty).$
We also defined a non-negative valued function $y = \cosh^{-1}(x)$:
$\{(x, y) | x = \frac{e^y + e^{-y}}{2}, y \ge 0\}.\quad\quad\quad(2)$
Simplifying $x=\frac{e^y+e^{-y}}{2}$ yields
$(e^y)^2-2x\cdot e^y+1=0.$
It follows that either $e^y = x-\sqrt{x^2-1}$ or $e^y=x+\sqrt{x^2-1}.$
For both expressions’ right-hand side to be valid requires that $x \ge 1$. However, when $x = 2$,
$e^y = x-\sqrt{x^2-1} = 2 - \sqrt{3} < 1$
suggests that $y < 0$ (see Exercise-2,3), contradicts (2).
Hence,
$e^y = x+\sqrt{x^2-1} \implies y = \log(x+\sqrt{x^2-1}).$
i.e.,
$\cosh^{-1}(x) = \log(x+\sqrt{x^2-1}), \;\;x \in [1, +\infty).$
Exercise-1 Show that $\forall x \in R, x-\sqrt{x^2+1} < 0.$
Exercise-2 Without calculator or CAS, show that $2-\sqrt{3} < 1.$
Exercise-3 Prove $\forall t \ge 0. e^t \ge 1$ (hint: see “Two Peas in a Pod, Part 2“) |
Question Video: Identifying and Proving Lines Parallel Using Angle Relationships | Nagwa Question Video: Identifying and Proving Lines Parallel Using Angle Relationships | Nagwa
# Question Video: Identifying and Proving Lines Parallel Using Angle Relationships Mathematics • First Year of Preparatory School
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Are the lines ๐ฟโ and ๐ฟโ
parallel?
03:06
### Video Transcript
Are lines ๐ฟ one and ๐ฟ five parallel?
๐ฟ five is the blue line. And ๐ฟ one is the red line. To prove these lines are parallel, we need to look for transversals. A transversal is a line that cuts across two other lines. We need a line that cuts through both the red and the blue line. ๐ฟ two, the pink line, is a transversal. ๐ฟ four, the green line, is a transversal. And the purple line ๐ฟ three is also a transversal.
However, we only need one of these to help us prove that ๐ฟ one and ๐ฟ five are parallel. So letโs consider the pink line, ๐ฟ two. In order for ๐ฟ one and ๐ฟ five to be parallel, the transversals have to create corresponding congruent angles.
Is there a way we can find out what this angle is? Well, we know that these three angles together make up a straight line. And that means if we subtract 45 and 80 from 180 degrees, we can find the missing angle. 180 degrees minus 45 degrees minus 80 degrees equals 55 degrees. The angle across from that 55 degrees is a vertical angle and also must measure 55 degrees.
If we look at the angles created with ๐ฟ five and ๐ฟ two in the blue line, we also see a pair of vertical angles that measure 55 degrees, these two corresponding angles. These two pair of corresponding congruent angles prove that ๐ฟ one and ๐ฟ five are parallel. ๐ฟ one and ๐ฟ five are parallel by corresponding congruent angles. Theyโre angles that are at the same corner at each intersection.
Itโs a little bit hard to see. So Iโm gonna remove the other two transversals since we only need one to prove these lines are parallel. Okay, it looks like this. When two lines intersect, they form four angles.
If we labelled them like this at the intersection of ๐ฟ one and ๐ฟ two, we would label them in the same way at the intersection of ๐ฟ two and ๐ฟ five. Weโre seeing that for top left angle, here the one labelled two, is 55 degrees for both ๐ฟ five intersection and ๐ฟ one intersection. Thatโs what we mean by corresponding. And the fact that theyโre both equal to 55 degrees makes them congruent angles.
Yes, by corresponding congruent angles, ๐ฟ one and ๐ฟ five are parallel.
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# Evaluate the double integral y^2 dA, D is the triangular region with vertices (0, 1), (1,2), (4,1)
This article aims to find the double integral of the triangular region with vertices. This article uses the concept of double integration. The definite integral of a positive function of one variable represents the area of the region between graph of the function and the $x-axis$. Similarly, the double integral of a positive function of two variables represents the volume of the region between the defined surface function (on the three-dimensional Cartesian plane, where $z = f(x, y)$ ) and the plane that contains its domain.
The points are:
$P (0,1) , Q(1,2) \: and \: R(4,1)$
The equation of line between $P$ and $R$ is given as:
$y = 1$
The equation of line between $P$ and $Q$ is given as:
Slope-intercept equation is given as:
$y = mx +c$
The slope is:
$m_{1} = \dfrac{2-1}{1-0} =1$
$m_{1} = 1$
and the line is passing over the point:
$x = 0\: , y = 1$
$1 = 0+ b$
$b = 1$
$y = x+1$
$x = y-1$
The equation for the line between $Q$ and $R$ is:
$m_{2} = \dfrac{(1-2)}{(4-1)} = -\dfrac{1}{3}$
$y = (-\dfrac{1}{3}) \times x +b$
$x =1 , y =2$
$2 = (-\dfrac{1}{3}) \times 1+ b$
$b = \dfrac{7}{3}$
$y = -(\dfrac{x}{3})+ \dfrac{7}{3}$
$x = 7 -3y$
The double integral becomes:
$A = \int \int y^{2} dx dy$
$A = \int y^{2} dy\int dx$
$A = \int y^{2} dy\int dx$
$A = \int_{1}^{2} y^{2} dy \times x |_{(y-1)}^{(7-3y)}$
$A = \int_{1}^{2} y^{2} dy \times (7-3y) – (y-1)$
$A = \int_{1}^{2} y^{2} dy \times (8-4y )$
$A = \int_{1}^{2} (8 y^{2} -4y^{3}) dy$
$= (\dfrac{8}{3} y^{3} – y^{4}) |_{1}^{2}$
$= \dfrac{56}{3} -15$
$A = \dfrac{11}{3}$
## Numerical Result
The solution is $A = \dfrac{11}{3}\: square\:units$.
## Example
Evaluate the double integral. $4 y^{2}\: dA$, $D$ is a triangular region with vertices $(0, 1), (1, 2), (4, 1)$.
Solution
The points are:
$P (0,1) , Q(1,2) \: and \: R(4,1)$
The equation of line between $P$ and $R$ is given as:
$y = 1$
The equation of line between $P$ and $Q$ is given as:
Slope-intercept equation is given as:
$y = mx +c$
The slope is:
$m_{1} = \dfrac{2-1}{1-0} =1$
$m_{1} = 1$
and the line is passing over the point:
$x = 0\: , y = 1$
$1 = 0+ b$
$b = 1$
$y = x+1$
$x = y-1$
The equation for the line between $Q$ and $R$ is:
$m_{2} = \dfrac{(1-2)}{(4-1)} = -\dfrac{1}{3}$
$y = (-\dfrac{1}{3}) \times x +b$
$x =1 , y =2$
$2 = (-\dfrac{1}{3}) \times 1+ b$
$b = \dfrac{7}{3}$
$y = -(\dfrac{x}{3})+ \dfrac{7}{3}$
$x = 7 -3y$
The double integral becomes:
$A = 4\int \int y^{2} dx dy$
$A = 4\int y^{2} dy\int dx$
$A = 4\int y^{2} dy\int dx$
$A = 4\int_{1}^{2} y^{2} dy \times x |_{(y-1)}^{(7-3y)}$
$A = 4\int_{1}^{2} y^{2} dy \times (7-3y) – (y-1)$
$A = 4\int_{1}^{2} y^{2} dy \times (8-4y )$
$A = 4\int_{1}^{2} (8 y^{2} -4y^{3}) dy$
$= 4(\dfrac{8}{3} y^{3} – y^{4}) |_{1}^{2}$
$=4(\dfrac{56}{3} -15)$
$A = 4(\dfrac{11}{3})$
$A = \dfrac{44}{3}$
The solution is $A = \dfrac{44}{3}\: square\:units$. |
# What is the integral of cos^5(x)?
Dec 16, 2014
$= \sin x + \frac{{\sin}^{5} x}{5} - \frac{2}{3} \cdot {\sin}^{3} x + c$, where $c$ is a constant
Explanation :
$= \int \left({\cos}^{5} x\right) \mathrm{dx}$
From trigonometric identity, which is
${\cos}^{2} x + {\sin}^{2} x = 1$, $\implies {\cos}^{2} x = 1 - \sin 2 x$
$= \int \left({\cos}^{4} x\right) \cdot \cos \left(x\right) \mathrm{dx}$
$= \int {\left({\cos}^{2} x\right)}^{2} \cdot \cos \left(x\right) \mathrm{dx}$
$= \int {\left(1 - {\sin}^{2} x\right)}^{2} \cdot \cos \left(x\right) \mathrm{dx}$ .. $\left(i\right)$
let's assume $\sin x = t$, $\implies \left(\cos x\right) \mathrm{dx} = \mathrm{dt}$
substituting this in the $\left(i\right)$, we get
$= \int {\left(1 - {t}^{2}\right)}^{2} \mathrm{dt}$
Now using expansion of ${\left(1 - y\right)}^{2} = 1 + {y}^{2} - 2 y$, yields,
$= \int \left(1 + {t}^{4} - 2 {t}^{2}\right) \mathrm{dt}$
$= \int \mathrm{dt} + \int {t}^{4} \mathrm{dt} - 2 \int {t}^{2} \mathrm{dt}$
$= t + {t}^{5} / 5 - 2 \cdot {t}^{3} / 3 + c$, where $c$ is a constant
$= t + {t}^{5} / 5 - \frac{2}{3} \cdot {t}^{3} + c$, where $c$ is a constant
now substituting $t$ back gives,
$= \sin x + {\left(\sin x\right)}^{5} / 5 - \frac{2}{3} \cdot {\left(\sin x\right)}^{3} + c$, where $c$ is a constant
$= \sin x + \frac{{\sin}^{5} x}{5} - \frac{2}{3} \cdot {\sin}^{3} x + c$, where $c$ is a constant |
# Online Statistics Education B - Online Statistics
```14. Regression
A. Introduction to Simple Linear Regression
B. Partitioning Sums of Squares
C. Standard Error of the Estimate
D. Inferential Statistics for b and r
E. Influential Observations
F. Regression Toward the Mean
G. Introduction to Multiple Regression
H. Exercises
This chapter is about prediction. Statisticians are often called upon to develop
methods to predict one variable from other variables. For example, one might want
to predict college grade point average from high school grade point average. Or,
one might want to predict income from the number of years of education.
462
Introduction to Linear Regression
by David M. Lane
Prerequisites
• Chapter 3: Measures of Variability
• Chapter 4: Describing Bivariate Data
Learning Objectives
1. Define linear regression
2. Identify errors of prediction in a scatter plot with a regression line
In simple linear regression, we predict scores on one variable from the scores on a
second variable. The variable we are predicting is called the criterion variable and
is referred to as Y. The variable we are basing our predictions on is called the
predictor variable and is referred to as X. When there is only one predictor
variable, the prediction method is called simple regression. In simple linear
regression, the topic of this section, the predictions of Y when plotted as a function
of X form a straight line.
The example data in Table 1 are plotted in Figure 1. You can see that there is
a positive relationship between X and Y. If you were going to predict Y from X, the
higher the value of X, the higher your prediction of Y.
Table 1. Example data.
X
Y
1.00
1.00
2.00
2.00
3.00
1.30
4.00
3.75
5.00
2.25
463
5
4
Y
3
2
1
0
0
1
2
3
4
5
6
X
Figure 1. A scatter plot of the example data.
Linear regression consists of finding the best-fitting straight line through the points.
The best-fitting line is called a regression line. The black diagonal line in Figure 2
is the regression line and consists of the predicted score on Y for each possible
value of X. The vertical lines from the points to the regression line represent the
errors of prediction. As you can see, the red point is very near the regression line;
its error of prediction is small. By contrast, the yellow point is much higher than
the regression line and therefore its error of prediction is large.
464
5
4
3.75
Y
3
2.25
2
2.00
1.30
1
1.00
0
0
1
2
3
4
5
6
X
Figure 2. A scatter plot of the example data. The black line consists of the
predictions, the points are the actual data, and the vertical lines
between the points and the black line represent errors of prediction.
The error of prediction for a point is the value of the point minus the predicted
value (the value on the line). Table 2 shows the predicted values (Y') and the errors
of prediction (Y-Y'). For example, the first point has a Y of 1.00 and a predicted Y
of 1.21. Therefore, its error of prediction is -0.21.
Table 2. Example data.
X
Y
Y'
Y-Y'
(Y-Y')2
1.00
1.00
1.210
-0.210
0.044
2.00
2.00
1.635
0.365
0.133
3.00
1.30
2.060
-0.760
0.578
4.00
3.75
2.485
1.265
1.600
5.00
2.25
2.910
-0.660
0.436
You may have noticed that we did not specify what is meant by “best-fitting line.”
By far the most commonly used criterion for the best-fitting line is the line that
minimizes the sum of the squared errors of prediction. That is the criterion that was
465
used to find the line in Figure 2. The last column in Table 2 shows the squared
errors of prediction. The sum of the squared errors of prediction shown in Table 2
is lower than it would be for any other regression line.
The formula for a regression line is
Y' = bX + A
where Y' is the predicted score, b is the slope of the line, and A is the Y intercept.
The equation for the line in Figure 2 is
Y' = 0.425X + 0.785
For X = 1,
Y' = (0.425)(1) + 0.785 = 1.21.
For X = 2,
Y' = (0.425)(2) + 0.785 = 1.64.
Computing the Regression Line
In the age of computers, the regression line is typically computed with statistical
software. However, the calculations are relatively easy are given here for anyone
who is interested. The calculations are based on the statistics shown in Table 3. MX
is the mean of X, MY is the mean of Y, sX is the standard deviation of X, sY is the
standard deviation of Y, and r is the correlation between X and Y.
Table 3. Statistics for computing the regression line
MX
3
on to linear regression
MY
sX
sY
r
2.06
1.581
1.072
0.627
The slope (b) can be calculated as follows:
=
and the intercept (A) can be calculated as
= (0.627)
1.072
= 0.425
1.581
466
roduction to linear regression
=
A = MY - bMX.
For these data,
= (0.627)
1.072
= 0.425
1.581
A = 2.06 - (0.425)(3) = 0.785
rtitioning the sums of squares
Note that the calculations have all been shown in terms of sample statistics rather
than population parameters. The formulas are the same; simply use the parameter
values for means, standard deviations, and the correlation.
=
(
Standardized Variables
)
The regression equation is simpler if variables are standardized so that their means
are equal to 0 and standard deviations are equal to 1, for then b = r and A = 0. This
makes the regression line:
=
(
)
ZY' = (r)(ZX)
where ZY' is the predicted standard score for Y, r is the correlation, and ZX is the
standardized score for X. Note that the slope of the regression equation for
=
standardized variables- is r.
Figure 3 shows a scatterplot with the regression line predicting the
standardized Verbal SAT from the standardized Math SAT.
A Real Example
-
-
=
The case study, “SAT and College GPA” contains high school and university
grades for 105 computer science majors at a local state school. We now consider
how we could predict a student's university GPA if we knew his or her high school
GPA.
=
+
Figure 3 shows a scatter plot of University GPA as a function of High School
GPA. You can see from the figure that there is a strong positive relationship. The
correlation is 0.78. The regression equation is
Univ GPA' = (0.675)(High School GPA) + 1.097
467
Therefore, a student with a high school GPA of 3 would be predicted to have a
university GPA of
University GPA' = (0.675)(3) + 1.097 = 3.12.
3.8
3.6
3.4
Univ_GPA
3.2
3
2.8
2.6
2.4
2.2
2
2
2.2
2.4
2.6
2.8
3
3.2
3.4
3.6
3.8
High_GPA
Figure 3. University GPA as a function of High School GPA.
Assumptions
It may surprise you, but the calculations shown in this section are assumption free.
Of course, if the relationship between X and Y is not linear, a different shaped
function could fit the data better. Inferential statistics in regression are based on
several assumptions, and these assumptions are presented in a later section of this
chapter.
468
Partitioning the Sums of Squares
by David M. Lane
duction to Prerequisites
linear regression
• Chapter 14: Introduction to Linear Regression
Learning Objectives
1. Compute the sum
= of squares Y
2. Convert raw scores to deviation scores
3. Compute predicted scores from a regression equation
4. Partition sum of squares Y into sum of squares predicted and sum of squares
1.072
error
(
)
=
0.627
= 0.425
5. Define r2 in terms
of sum of squares explained and sum of squares Y
1.581
One useful aspect of regression is that it can divide the variation in Y into two
parts: the variation of the predicted scores and the variation in the errors of
variation of Y is called the sum of squares Y and is defined as the
ioning theprediction.
sums ofThe
squares
sum of the squared deviations of Y from the mean of Y. In the population, the
formula is
=
)
(
where SSY is the sum of squares Y, Y is an individual value of Y, and my is the
mean of Y. A simple example is given in Table 1. The mean of Y is 2.06 and SSY is
)
=
(
the sum of the values in the third column and is equal to 4.597.
Table 1. Example of SSY.
Y
Y-my
(Y-my)2
1.00
- -1.06
=1.1236
2.00
-0.06
0.0036
1.30
-0.76
0.5776
1.69
-
2.8561
0.19
0.0361
3.75
2.25
-
=
=
+
469
=
)
(
When computed in a sample, you should use the sample mean, M, in place of the
population mean:
=
)
(
It is sometimes convenient to use formulas that use deviation scores rather than
raw scores. Deviation scores are simply deviations from the mean. By convention,
small letters rather
- than capitals=are used for deviation scores. Therefore, the score,
y indicates the difference between Y and the mean of Y. Table 2 shows the use of
this notation. The numbers are the same as in Table 1.
Table 2. Example of SSY using Deviation Scores.
-
-
=
Y
y
y2
1.00
-1.06
1.1236
-0.06
+
0.0036
1.30
-0.76
0.5776
3.75
1.69
2.8561
2.25
0.19
0.0361
10.30
0.00
4.5970
2.00
=
The data in Table 3 are reproduced from the introductory section. The column X
has the values of the predictor variable and the column Y has the criterion
variable. The third column, y, contains the the differences between the column Y
and the mean of Y.
470
Table 3. Example data. The last row contains column sums.
X
Y
y
y2
Y'
y'
y'2
Y-Y'
(Y-Y')2
1.00
1.00
-1.06
1.1236
1.210
-0.850
0.7225
-0.210
0.044
2.00
2.00
-0.06
0.0036
1.635
-0.425
0.1806
0.365
0.133
3.00
1.30
-0.76
0.5776
2.060
0.000
0.000
-0.760
0.578
4.00
3.75
1.69
2.8561
2.485
0.425
0.1806
1.265
1.600
5.00
2.25
0.19
0.0361
2.910
0.850
0.7225
-0.660
0.436
15.00
10.30
0.00
4.597
10.300
0.000
1.806
0.000
2.791
The fourth column, y2, is simply the square of the y column. The column Y'
contains the predicted values of Y. In the introductory section, it was shown that
the equation for the regression line for these data is
Y' = 0.425X + 0.785.
The values of Y' were computed according to this equation. The column y' contains
deviations of Y' from the mean of Y' and y'2 is the square of this column. The nextto-last column, Y-Y', contains the actual scores (Y) minus the predicted scores (Y').
The last column contains the squares of these errors of prediction.
We are now in a position to see how the SSY is partitioned. Recall that SSY
is the sum of the squared deviations from the mean. It is therefore the sum of the y2
column and is equal to 4.597. SSY can be partitioned into two parts: the sum of
squares predicted (SSY') and the sum of squares error (SSE). The sum of squares
predicted is the sum of the squared deviations of the predicted scores from the
mean predicted score. In other words, it is the sum of the y'2 column and is equal to
1.806. The sum of squares error is the sum of the squared errors of prediction. It is
therefore the sum of the (Y-Y')2 column and is equal to 2.791. This can be summed
up as:
SSY = SSY' + SSE
4.597 = 1.806 + 2.791
471
1.072
= (0.627)
= 0.425
1.072
(0.6271.581
)
==
= 0.425
1.581
Partitioning the sums of squares
Therethe
aresums
severalofother
notable features about Table 3. First, notice that the sum of
Partitioning
squares
y and=
the(0.627
sum of)y'1.072
are both
This will always be the case because these
= zero.
0.425
1.581
variables were created
by subtracting their respective means from each value. Also,
= of Y-Y'
( is 0.) This indicates that although some Y values are
notice that the mean
)
=
(
higher than their respective predicted Y values and some are lower, the average
difference
zero.
oning the sums
of issquares
The SSY is the
the SSY' is the variation explained, and the
)
= total
( variation,
SSE is the variation unexplained.
Therefore,
the proportion of variation explained
)
=
(
can be computed as:
=
)
(
-
=
-
=
Similarly, the proportion not explained is:
=
(
-
)
-
-
-
=
=
+
There is an important =
relationship
between the proportion of variation explained
=+
2
=
and Pearson's correlation: r is the proportion of variation explained. Therefore, if r
= 1, then, naturally, the proportion of variation explained is 1; if r = 0, then the
proportion explained is 0. One last example: for r = 0.4, the proportion of variation
explained is 0.16.
- variance
= by dividing the variation by N (for a
Since the
is computed
population) or N-1 (for a sample), the relationships spelled out above in terms of
variation also hold for variance. For example,
=
+
where the first term is the variance total, the second term is the variance of Y', and
the last term is the variance of the errors of prediction (Y-Y'). Similarly, r2 is the
proportion of variance explained as well as the proportion of variation explained.
Summary Table
It is often convenient to summarize the partitioning of the data in a table such as
Table 4. The degrees of freedom column (df) shows the degrees of freedom for
each source of variation. The degrees of freedom for the sum of squares explained
472
is equal to the number of predictor variables. This will always be 1 in simple
regression. The error degrees of freedom is equal to the total number of
observations minus 2. In this example, it is 5 - 2 = 3. The total degrees of freedom
is the total number of observations minus 1.
Table 4. Summary Table for Example Data
Source
Sum of Squares
df
Mean Square
Explained
1.806
1
1.806
Error
2.791
3
0.930
Total
4.597
4
473
Standard Error of the Estimate
by David M. Lane
Prerequisites
• Chapter 3: Measures of Variability
• Chapter 14: Introduction to Linear Regression
• Chapter 14: Partitioning Sums of Squares
Learning Objectives
1. Make judgments about the size of the standard error of the estimate from a
scatter plot
2. Compute the standard error of the estimate based on errors of prediction
3. Compute the standard error using Pearson's correlation
4. Estimate the standard error of the estimate based on a sample
Figure 1 shows two regression examples. You can see that in Graph A, the points
are closer to the line than they are in Graph B. Therefore, the predictions in Graph
A are more accurate than in Graph B.
60
60
A
40
40
30
30
Y2
50
Y1
50
B
20
20
10
10
0
0
-10
-10
20
30
40
50
X
60
70
80
20
30
40
50
X
60
70
80
Figure 1. Regressions differing in accuracy of prediction.
The standard error of the estimate is a measure of the accuracy of predictions.
Recall that the regression line is the line that minimizes the sum of squared
deviations of prediction (also called the sum of squares error). The standard error
of the estimate is closely related to this quantity and is defined below:
474
where σest is the standard error of the estimate, Y is an actual score, Y' is a
predicted score, and N is the number of pairs of scores. The numerator is the sum
of squared differences between the actual scores and the predicted scores.
Note the similarity of the formula for σest to the formula for σ:
In fact, σest is the standard deviation of the errors of prediction (each Y - Y’ is an
error of prediction).
Assume the data in Table 1 are the data from a population of five X, Y pairs.
Table 1. Example data.
X
Y
Y'
Y-Y'
(Y-Y')2
1.00
1.00
1.210
-0.210
0.044
2.00
2.00
1.635
0.365
0.133
3.00
1.30
2.060
-0.760
0.578
4.00
3.75
2.485
1.265
1.600
2.910
-0.660
0.436
10.30
0.000
2.791
ndard error of the Estimate
5.00
Sum
(
=
15.00
2.25
)
10.30
The last column shows that the sum of the squared errors of prediction is 2.791.
Therefore, the standard error of the estimate is
=
2.791
= 0.747
5
There is a version of the formula for the standard error in terms of Pearson's
correlation:
=
(1
)
475
=
(
)
= 2.791 =
= 0.747
0.7472.791
=
=
= 0.747
5
)
5(
5
=
(1
))
(1
(1
=
= 2.791
=
=
= 0.747
5
)
where ρ is the population value of Pearson's correlation and SSY is
=
=
((
(1
=
))=
(
)
)
For the data in Table 1, my = 10.30, SSY = 4.597 and r = 0.6268. Therefore,
(11
(
=
=
0.6268
0.6268
5
5=
=
2.791
(4.597)
))(4.597)
2.791
=
= 0.747
0.7472.791
(1 0.6268
)(4.597)
=
=
5
=
= 0.747
5
5 5
(
)
which is the same value computed previously.
Similar formulas( are used) when the standard error of the estimate is
)
(
=a sample
) The only difference is that the
(
computed from=
rather
than a population.
2
(1 0.6268 )(4.597) 2 =2.791
N-2
2 is used rather than N-1 is that
= denominator is N-2 rather than
= N. The reason
= 0.747
5
5
two parameters (the slope and the intercept) were estimated in order to estimate the
sum of squares. Formulas for a sample comparable to the ones for a population are
shown below:
=
)
(
2
2.791
= 0.964
3
2.791
=
= 0.964
3
=
=
=
(1
(1
ential statistics for b and r
rential statistics for b and r
)
2)
2
476
Inferential Statistics for b and r
by David M. Lane
Prerequisites
• Chapter 9: Sampling Distribution of r
• Chapter 9: Confidence Interval for r
Learning Objectives
1. State the assumptions that inferential statistics in regression are based upon
2. Identify heteroscedasticity in a scatter plot
3. Compute the standard error of a slope
4. Test a slope for significance
5. Construct a confidence interval on a slope
6. Test a correlation for significance
7. Construct a confidence interval on a correlation
This section shows how to conduct significance tests and compute confidence
intervals for the regression slope and Pearson's correlation. As you will see, if the
regression slope is significantly different from zero, then the correlation coefficient
is also significantly different from zero.
Assumptions
Although no assumptions were needed to determine the best-fitting straight line,
assumptions are made in the calculation of inferential statistics. Naturally, these
assumptions refer to the population, not the sample.
1. Linearity: The relationship between the two variables is linear.
2. Homoscedasticity: The variance around the regression line is the same for all
values of X. A clear violation of this assumption is shown in Figure 1. Notice
that the predictions for students with high high-school GPAs are very good,
whereas the predictions for students with low high-school GPAs are not very
good. In other words, the points for students with high high-school GPAs are
close to the regression line, whereas the points for low high-school GPA
students are not.
477
3.8
3.6
3.4
Univ_GPA
3.2
3
2.8
2.6
2.4
2.2
=
2
2
2.2
2.4
2.6
2.791
= 0.964
3
2.8
3
3.2
3.4
3.6
3.8
High_GPA
Figure 1. University GPA as a function of High School GPA.
=
)
2
(1
3. The errors of prediction are distributed normally. This means that the
distributions of deviations from the regression line are normally distributed. It does
not mean that X or Y is normally distributed.
Inferential statistics for b and r
Significance Test for the Slope (b)
Recall the general formula for a t test:
=
-
-
-
-
-
As applied here, the statistic is the sample value of the slope (b) and the
hypothesized value is 0. The degrees of freedom for this test are:
=
df = N-2
where N is the number of pairs of scores.
=
(
)
478
ential statistics for b and r
-
=
-
=
-
-
-
-
The estimated standard error of b is computed using the following formula:
-
-
-
-
-
=
=
where sb is the estimated standard error of b, sest is the standard error of the
estimate, and SSX is the sum of squared deviations of X from the mean of X. SSX
)
= as (
is calculated
)
= 2.791
(
=
= 0.964
0.964 3
=
= 0.305
where Mx is 10
the mean of X. As shown previously, the standard error of the estimate
can be calculated0.964
as
=
= 0.305
10
)
(1
=
2
These formulas are illustrated with the data shown in Table 1. These data are
reproduced
the rintroductory section. The column X has the values of the
ential statistics
for from
b and
predictor variable and the column Y has the values of the criterion variable. The
third column, x, contains the differences between the values of column X and the
mean of X. The fourth column, x2, is the square of the x column. The fifth column,
= y, contains the differences between the values of column Y and the mean of Y. The
- y2, is simply
- the square
- - of the
- y column.
last column,
=
=
=
(
0.964
10
)
= 0.305
479
)
2
(1
= data.
Table 1. Example
X
Y
x
x2
y
y2
1.00
1.00
-2.00
4
-1.06
1.1236
2.00
-1.00
1
-0.06
0.0036
3.00
1.30
0.00
0
-0.76
0.5776
4.00
3.75
1.00
1
1.69
2.8561
5.00
2.25
2.00
4
0.19
0.0361
10.00
0.00
4.5970
2.00 r
ntial statistics for b and
=
Sum
-
15.00
-
-
-
-
10.30
-
0.00
The computation of the standard error of the estimate (sest) for these data is shown
in the section on the standard error of the estimate. It is equal to 0.964.
sest = =
0.964
SSX is the sum of squared deviations from the mean of X. It is, therefore, equal to
the sum of the x2 column and is equal to 10.
)
(
SSX =
= 10.00
We now have all the information to compute the standard error of b:
=
0.964
10
= 0.305
As shown previously, the slope (b) is 0.425. Therefore,
=
0.425
= 1.39
0.305
df = N-2 = 5-2 = 3.
2
The p value for
= a two-tailed t test is 0.26. Therefore, the slope is not significantly
different from 0. 1
=
0.627 5
1
2
0.627
= 1.39
480
Confidence Interval for the Slope
The method for computing a confidence interval for the population slope is very
similar to methods for computing other confidence intervals. For the 95%
confidence interval, the formula is:
lower limit: b - (t.95)(sb)
upper limit: b + (t.95)(sb)
where t.95 is the value of t to use for the 95% confidence interval.
The values of t to be used in a confidence interval can be looked up in a table
of the t distribution. A small version of such a table is shown in Table 2. The first
column, df, stands for degrees of freedom.
Table 2. Abbreviated t table.
df
0.95
0.99
2
4.303
9.925
3
3.182
5.841
4
2.776
4.604
5
2.571
4.032
8
2.306
3.355
10
2.228
3.169
20
2.086
2.845
50
2.009
2.678
100
1.984
2.626
You can also use the “inverse t distribution” calculator (external link; requires
Java) to find the t values to use in a confidence interval.
Applying these formulas to the example data,
lower limit: 0.425 - (3.182)(0.305) = -0.55
upper limit: 0.425 + (3.182)(0.305) = 1.40
Significance Test for the Correlation
The formula for a significance test of Pearson's correlation is shown below:
481
=
0.4250.425
= = 1.39= 1.39
0.305
0.305
=
2
1
=
2
1
where N is the number of pairs of scores. For the example data,
=
0.6270.627
5 25 2
= 1.39= 1.39
=
1 0.627
1 0.627
Notice that this is the same t value obtained in the t test of b. As in that test, the
1 N-2
+ =1 5-2
degrees of freedom is
+ = 3.
= 0.5 = 0.5
1
1
1
1
3
3
tion to multiple
regression
oduction
to multiple
regression
=
=
=
1
=
1
482
Influential Observations
by David M. Lane
Prerequisites
• Chapter 14: Introduction to Linear Regression
Learning Objectives
1. Define “influence”
2. Describe what makes a point influential
3. Define “leverage”
4. Define “distance”
It is possible for a single observation to have a great influence on the results of a
regression analysis. It is therefore important to be alert to the possibility of
influential observations and to take them into consideration when interpreting the
results.
Influence
The influence of an observation can be thought of in terms of how much the
predicted scores for other observations would differ if the observation in question
were not included. Cook's D is a good measure of the influence of an observation
and is proportional to the sum of the squared differences between predictions made
with all observations in the analysis and predictions made leaving out the
observation in question. If the predictions are the same with or without the
observation in question, then the observation has no influence on the regression
model. If the predictions differ greatly when the observation is not included in the
analysis, then the observation is influential.
A common rule of thumb is that an observation with a value of Cook's D
over 1.0 has too much influence. As with all rules of thumb, this rule should be
applied judiciously and not thoughtlessly.
An observation's influence is a function of two factors: (1) how much the
observation's value on the predictor variable differs from the mean of the predictor
variable and (2) the difference between the predicted score for the observation and
its actual score. The former factor is called the observation's leverage. The latter
factor is called the observation's distance.
483
Calculation of Cook's D (Optional)
The first step in calculating the value of Cook's D for an observation is to predict
all the scores in the data once using a regression equation based on all the
observations and once using all the observations except the observation in
question. The second step is to compute the sum of the squared differences
between these two sets of predictions. The final step is to divide this result by 2
times the MSE (see the section on partitioning the variance).
Leverage
The leverage of an observation is based on how much the observation's value on
the predictor variable differs from the mean of the predictor variable. The greater
an observation's leverage, the more potential it has to be an influential observation.
For example, an observation with the mean on the predictor variable has no
influence on the slope of the regression line regardless of its value on the criterion
variable. On the other hand, an observation that is extreme on the predictor
variable has, depending on its distance, the potential to affect the slope greatly.
Calculation of Leverage (h)
The first step is to standardize the predictor variable so that it has a mean of 0 and a
standard deviation of 1. Then, the leverage (h) is computed by squaring the
observation's value on the standardized predictor variable, adding 1, and dividing
by the number of observations.
Distance
The distance of an observation is based on the error of prediction for the
observation: The greater the error of prediction, the greater the distance. The most
commonly used measure of distance is the studentized residual. The studentized
residual for an observation is closely related to the error of prediction for that
observation divided by the standard deviation of the errors of prediction. However,
the predicted score is derived from a regression equation in which the observation
in question is not counted. The details of the computation of a studentized residual
are a bit complex and are beyond the scope of this work.
An observation with a large distance will not have that much influence if its
leverage is low. It is the combination of an observation's leverage and distance that
determines its influence.
484
Example
Table 1 shows the leverage, studentized residual, and influence for each of the five
observations in a small dataset.
Table 1. Example Data.
ID
X
Y
h
R
D
A
1
2
0.39
-1.02
0.40
B
2
3
0.27
-0.56
0.06
C
3
5
0.21
0.89
0.11
D
4
6
0.20
1.22
0.19
E
8
7
0.73
-1.68
8.86
h is the leverage, R is the studentized residual, and D is Cook's measure of influence.
Observation A has fairly high leverage, a relatively high residual, and
moderately high influence.
Observation B has small leverage and a relatively small residual. It has very
little influence.
Observation C has small leverage and a relatively high residual. The
influence is relatively low.
Observation D has the lowest leverage and the second highest residual.
Although its residual is much higher than Observation A, its influence
is much less because of its low leverage.
Observation E has by far the largest leverage and the largest residual. This
combination of high leverage and high residual makes this
observation extremely influential.
Figure 1 shows the regression line for the whole dataset (blue) and the regression
line if the observation in question is not included (red) for all observations. The
observation in question is circled. Naturally, the regression line for the whole
dataset is the same in all panels. The residual is calculated relative to the line for
which the observation in question is not included in the analysis. This can be seen
most clearly for Observation E which lies very close to the regression line
485
computed when it is included but very far from the regression line when it is
excluded from the calculation of the line.
8
8
A
7
B
7
6
5
5
Y
Y
6
4
4
3
3
Leverage: 0.39
Residual: -1.02
Influence: 0.40
2
Leverage: 0.27
Residual: -0.56
Influence: 0.06
2
1
1
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
X
5
6
7
8
9
X
8
8
C
7
D
7
6
5
5
Y
Y
6
4
4
3
Leverage: 0.21
Residual: 0.89
Influence: 0.11
2
3
2
1
0
1
2
3
4
5
6
7
8
Leverage: 0.20
Residual: 1.22
Influence: 0.19
9
X
1
0
1
2
3
4
5
6
7
8
9
X
8
E
7
6
Y
5
4
3
Leverage: 0.73
Residual: -1.68
Influence: 8.86
2
1
0
1
2
3
4
5
6
7
8
9
X
Figure 1. Illustration of leverage, residual, and influence.
486
The circled points are not included in the calculation of the red regression line. All
points are included in the calculation of the blue regression line.
487
Regression Toward the Mean
by David M. Lane
Prerequisites
• Chapter 14: Regression Introduction
Learning Objectives
1. Explain what regression towards the mean is
2. State the conditions under which regression toward the mean occurs
3. Identify situations in which neglect of regression toward the mean leads to
incorrect conclusions
4. Explain how regression toward the mean relates to a regression equation.
Regression toward the mean involves outcomes that are at least partly due to
chance. We begin with an example of a task that is entirely chance: Imagine an
experiment in which a group of 25 people each predicted the outcomes of flips of a
fair coin. For each subject in the experiment, a coin is flipped 12 times and the
subject predicts the outcome of each flip. Figure 1 shows the results of a simulation
of this “experiment.” Although most subjects were correct from 5 to 8 times out of
12, one simulated subject was correct 10 times. Clearly, this subject was very lucky
and probably would not do as well if he or she performed the task a second time. In
fact, the best prediction of the number of times this subject would be correct on the
retest is 6 since the probability of being correct on a given trial is 0.5 and there are
12 trials.
488
7
6
5
4
3
2
1
0
2
3
4
5
6
7
8
9
10
11
Figure 1. Histogram of results of a simulated experiment.
More technically, the best prediction for the subject's result on the retest is the
mean of the binomial distribution with N = 12 and p = 0.50. This distribution is
shown in Figure 2 and has a mean of 6.
0.25
0.2
0.15
0.1
0.05
0
0
1
3
4
5
6
7
8
9
10
11
12
Figure 2. Binomial Distribution for N = 12 and p = .50.
The point here is that no matter how many coin flips a subject predicted correctly,
the best prediction of their score on a retest is 6.
489
Now we consider a test we will call “Test A” that is partly chance and partly
skill: Instead of predicting the outcomes of 12 coin flips, each subject predicts the
outcomes of 6 coin flips and answers 6 true/false questions about world history.
Assume that the mean score on the 6 history questions is 4. A subject's score on
Test A has a large chance component but also depends on history knowledge. If a
subject scored very high on this test (such as a score of 10/12), it is likely that they
did well on both the history questions and the coin flips. For example, if they only
got four of the history questions correct, they would have had to have gotten all six
of the coin predictions correct, and this would have required exceptionally good
luck. If given a second test (Test B) that also included coin predictions and history
questions, their knowledge of history would be helpful and they would again be
expected to score above the mean. However, since their high performance on the
coin portion of Test A would not be predictive of their coin performance on Test B,
they would not be expected to fare as well on Test B as on Test A. Therefore, the
best prediction of their score on Test B would be somewhere between their score
on Test A and the mean of Test B. This tendency of subjects with high values on a
measure that includes chance and skill to score closer to the mean on a retest is
called “regression toward the mean.”
The essence of the regression-toward-the-mean phenomenon is that people
with high scores tend to be above average in skill and in luck, and that only the
skill portion is relevant to future performance. Similarly, people with low scores
tend to be below average in skill and luck and their bad luck is not relevant to
future performance. This does not mean that all people who score high have above
average luck. However, on average they do.
Almost every measure of behavior has a chance and a skill component to it.
Take a student's grade on a final exam as an example. Certainly, the student's
knowledge of the subject will be a major determinant of his or her grade. However,
there are aspects of performance that are due to chance. The exam cannot cover
everything in the course and therefore must represent a subset of the material.
Maybe the student was lucky in that the one aspect of the course the student did
not understand well was not well represented on the test. Or, maybe, the student
was not sure which of two approaches to a problem would be better but, more or
less by chance, chose the right one. Other chance elements come into play as well.
Perhaps the student was awakened early in the morning by a random phone call,
490
resulting in fatigue and lower performance. And, of course, guessing on multiple
choice questions is another source of randomness in test scores.
There will be regression toward the mean in a test-retest situation whenever
there is less than a perfect (r = 1) relationship between the test and the retest. This
follows from the formula for a regression line with standardized variables shown
below.
ZY' = (r)(ZX)
From this equation it is clear that if the absolute value of r is less than 1, then the
predicted value of ZY will be closer to 0, the mean for standardized scores, than is
ZX. Also, note that if the correlation between X and Y is 0, as it would be for a task
that is all luck, the predicted standard score for Y is its mean, 0, regardless of the
score on X.
Figure 3 shows a scatter plot with the regression line predicting the
standardized Verbal SAT from the standardized Math SAT. Note that the slope of
the line is equal to the correlation of 0.835 between these variables.
491
2.5
2
Standardized Verbal SAT
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-2
-1.5
-1
-0.5
0
0.5
Standardized Math SAT
1
1.5
2
Figure 3. Prediction of Standardized Verbal SAT from Standardized Math
SAT.
The point represented by a blue diamond has a value of 1.6 on the standardized
Math SAT. This means that this student scored 1.6 standard deviations above the
mean on Math SAT. The predicted score is (r)(1.6) = (0.835)(1.6) = 1.34. The
horizontal line on the graph shows the value of the predicted score. The key point
is that although this student scored 1.6 standard deviations above the mean on
Math SAT, he or she is only predicted to score 1.34 standard deviations above the
mean on Verbal SAT. Thus, the prediction is that the Verbal SAT score will be
closer to the mean of 0 than is the Math SAT score. Similarly, a student scoring far
below the mean on Math SAT will be predicted to score higher on Verbal SAT.
Regression toward the mean occurs in any situation in which observations
are selected on the basis of performance on a task that has a random component. If
you choose people on the basis of their performance on such a task, you will be
choosing people partly on the basis of their skill and partly on the basis of their
luck on the task. Since their luck cannot be expected to be maintained from trial to
trial, the best prediction of a person's performance on a second trial will be
492
somewhere between their performance on the first trial and the mean performance
on the first trial. The degree to which the score is expected to “regress toward the
mean” in this manner depends on the relative contributions of chance and skill to
the task: the greater the role of chance, the more the regression toward the mean.
Errors Resulting From Failure to Understand Regression Toward the
Mean
Failure to appreciate regression toward the mean is common and often leads to
incorrect interpretations and conclusions. One of the best examples is provided by
Nobel Laureate Daniel Kahneman in his autobiography (external link). Dr.
Kahneman was attempting to teach flight instructors that praise is more effective
than punishment. He was challenged by one of the instructors who relayed that in
his experience praising a cadet for executing a clean maneuver is typically
followed by a lesser performance, whereas screaming at a cadet for bad execution
is typically followed by improved performance. This, of course, is exactly what
would be expected based on regression toward the mean. A pilot's performance,
although based on considerable skill, will vary randomly from maneuver to
maneuver. When a pilot executes an extremely clean maneuver, it is likely that he
or she had a bit of luck in their favor in addition to their considerable skill. After
the praise but not because of it, the luck component will probably disappear and
the performance will be lower. Similarly, a poor performance is likely to be partly
due to bad luck. After the criticism but not because of it, the next performance will
likely be better. To drive this point home, Kahneman had each instructor perform a
task in which a coin was tossed at a target twice. He demonstrated that the
performance of those who had done the best the first time deteriorated, whereas the
performance of those who had done the worst improved.
Regression toward the mean is frequently present in sports performance. A
good example is provided by Schall and Smith (2000), who analyzed many aspects
of baseball statistics including the batting averages of players in 1998. They chose
the 10 players with the highest batting averages (BAs) in 1998 and checked to see
how well they did in 1999. According to what would be expected based on
regression toward the mean, these players should, on average, have lower batting
averages in 1999 than they did in 1998. As can be seen in Table 1, 7/10 of the
players had lower batting averages in 1999 than they did in 1998. Moreover, those
who had higher averages in 1999 were only slightly higher, whereas those who
493
were lower were much lower. The average decrease from 1998 to 1999 was 33
points. Even so, most of these players had excellent batting averages in 1999
indicating that skill was an important component of their 1998 averages.
Table 1. How the Ten Players with the Highest BAs in 1998 did in 1999.
1998
1999
Difference
363
354
339
337
336
331
328
328
327
327
379
298
342
281
249
298
297
303
257
332
16
-56
3
-56
-87
-33
-31
-25
-70
5
Figure 4 shows the batting averages of the two years. The decline from 1998 to
1999 is clear. Note that although the mean decreased from 1998, some players
increased their batting averages. This illustrates that regression toward the mean
does not occur for every individual. Although the predicted scores for every
individual will be lower, some of the predictions will be wrong.
494
0.4
Batting Average
0.35
0.3
0.25
1998
Year
1999
Figure 4. Quantile plots of the batting averages. The line connects the means
of the plots.
Regression toward the mean plays a role in the so-called “Sophomore Slump,” a
good example of which is that a player who wins “rookie of the year” typically
does less well in his second season. A related phenomenon is called the Sports
Illustrated Cover Jinx.
An experiment without a control group can confound regression effects with
real effects. For example, consider a hypothetical experiment to evaluate a readingimprovement program. All first graders in a school district were given a reading
achievement test and the 50 lowest-scoring readers were enrolled in the program.
The students were retested following the program and the mean improvement was
large. Does this necessarily mean the program was effective? No, it could be that
the initial poor performance of the students was due, in part, to bad luck. Their luck
would be expected to improve in the retest, which would increase their scores with
or without the treatment program.
For a real example, consider an experiment that sought to determine whether
the drug propranolol would increase the SAT scores of students thought to have
495
test anxiety (external link). Propranolol was given to 25 high-school students
chosen because IQ tests and other academic performance indicated that they had
not done as well as expected on the SAT. On a retest taken after receiving
propranolol, students improved their SAT scores an average of 120 points. This
was a significantly greater increase than the 38 points expected simply on the basis
of having taken the test before. The problem with the study is that the method of
selecting students likely resulted in a disproportionate number of students who had
bad luck when they first took the SAT. Consequently, these students would likely
have increased their scores on a retest with or without the propranolol. This is not
to say that propranolol had no effect. However, since possible propranolol effects
and regression effects were confounded, no firm conclusions should be drawn.
Randomly assigning students to either the propranolol group or a control
group would have improved the experimental design. Since the regression effects
would then not have been systematically different for the two groups, a significant
difference would have provided good evidence for a propranolol effect.
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Introduction to Multiple Regression
by David M. Lane
Prerequisites
• Chapter 14: Simple Linear Regression
• Chapter 14: Partitioning Sums of Squares
• Chapter 14: Standard Error of the Estimate
• Chapter 14: Inferential Statistics for b and r
Learning Objectives
1. State the regression equation
2. Define “regression coefficient”
3. Define “beta weight”
4. Explain what R is and how it is related to r
5. Explain why a regression weight is called a “partial slope”
6. Explain why the sum of squares explained in a multiple regression model is
usually less than the sum of the sums of squares in simple regression
7. Define R2 in terms of proportion explained
8. Test R2 for significance
9. Test the difference between a complete and reduced model for significance
10. State the assumptions of multiple regression and specify which aspects of the
analysis require assumptions
In simple linear regression, a criterion variable is predicted from one predictor
variable. In multiple regression, the criterion is predicted by two or more variables.
For example, in the SAT case study, you might want to predict a student's
university grade point average on the basis of their High-School GPA (HSGPA)
and their total SAT score (verbal + math). The basic idea is to find a linear
combination of HSGPA and SAT that best predicts University GPA (UGPA). That
is, the problem is to find the values of b1 and b2 in the equation shown below that
gives the best predictions of UGPA. As in the case of simple linear regression, we
define the best predictions as the predictions that minimize the squared errors of
prediction.
UGPA' = b1HSGPA + b2SAT + A
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where UGPA' is the predicted value of University GPA and A is a constant. For
these data, the best prediction equation is shown below:
UGPA' = 0.541 x HSGPA + 0.008 x SAT + 0.540
In other words, to compute the prediction of a student's University GPA, you add
up (a) their High-School GPA multiplied by 0.541, (b) their SAT multiplied by
0.008, and (c) 0.540. Table 1 shows the data and predictions for the first five
students in the dataset.
Table 1. Data and Predictions.
HSGPA
SAT
UGPA'
3.45
1232
3.38
2.78
1070
2.89
2.52
1086
2.76
3.67
1287
3.55
3.24
1130
3.19
The values of b (b1 and b2) are sometimes called “regression coefficients” and
sometimes called “regression weights.” These two terms are synonymous.
The multiple correlation (R) is equal to the correlation between the predicted
scores and the actual scores. In this example, it is the correlation between UGPA'
and UGPA, which turns out to be 0.79. That is, R = 0.79. Note that R will never be
negative since if there are negative correlations between the predictor variables and
the criterion, the regression weights will be negative so that the correlation
between the predicted and actual scores will be positive.
Interpretation of Regression Coefficients
A regression coefficient in multiple regression is the slope of the linear relationship
between the criterion variable and the part of a predictor variable that is
independent of all other predictor variables. In this example, the regression
coefficient for HSGPA can be computed by first predicting HSGPA from SAT and
saving the errors of prediction (the differences between HSGPA and HSGPA').
These errors of prediction are called “residuals” since they are what is left over in
HSGPA after the predictions from SAT are subtracted, and they represent the part
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of HSGPA that is independent of SAT. These residuals are referred to as
HSGPA.SAT, which means they are the residuals in HSGPA after having been
predicted by SAT. The correlation between HSGPA.SAT and SAT is necessarily 0.
The final step in computing the regression coefficient is to find the slope of
the relationship between these residuals and UGPA. This slope is the regression
coefficient for HSGPA. The following equation is used to predict HSGPA from
SAT:
HSGPA' = -1.314 + 0.0036 x SAT
The residuals are then computed as:
HSGPA - HSGPA'
The linear regression equation for the prediction of UGPA by the residuals is
UGPA' = 0.541 x HSGPA.SAT + 3.173
Notice that the slope (0.541) is the same value given previously for b1 in the
multiple regression equation.
This means that the regression coefficient for HSGPA is the slope of the
relationship between the criterion variable and the part of HSPGA that is
independent of (uncorrelated with) the other predictor variables. It represents the
change in the criterion variable associated with a change of one in the predictor
variable when all other predictor variables are held constant. Since the regression
coefficient for HSGPA is 0.54, this means that, holding SAT constant, a change of
one in HSGPA is associated with a change of 0.54 in UGPA. If two students had
the same SAT and differed in HSGPA by 2, then you would predict they would
differ in UGPA by (2)(0.54) = 1.08. Similarly, if they differed by 0.5, then you
would predict they would differ by (0.50)(0.54) = 0.27.
The slope of the relationship between the part of a predictor variable
independent of other predictor variables and the criterion is its partial slope. Thus
the regression coefficient of 0.541 for HSGPA and the regression coefficient of
0.008 for SAT are partial slopes. Each partial slope represents the relationship
between the predictor variable and the criterion holding constant all of the other
predictor variables.
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It is difficult to compare the coefficients for different variables directly
because they are measured on different scales. A difference of 1 in HSGPA is a
fairly large difference, whereas a difference of 1 on the SAT is negligible.
Therefore, it can be advantageous to transform the variables so that they are on the
same scale. The most straightforward approach is to standardize the variables so
that they all have a standard deviation of 1. A regression weight for standardized
variables is called a “beta weight” and is designated by the Greek letter β. For
these data, the beta weights are 0.625 and 0.198. These values represent the change
in the criterion (in standard deviations) associated with a change of one standard
deviation on a predictor [holding constant the value(s) on the other predictor(s)].
Clearly, a change of one standard deviation on HSGPA is associated with a larger
difference than a change of one standard deviation of SAT. In practical terms, this
means that if you know a student's HSGPA, knowing the student's SAT does not
aid the prediction of UGPA much. However, if you do not know the student's
HSGPA, his or her SAT can aid in the prediction since the β weight in the simple
regression predicting UGPA from SAT is 0.68. For comparison purposes, the β
weight in the simple regression predicting UGPA from HSGPA is 0.78. As is
typically the case, the partial slopes are smaller than the slopes in simple
regression.
Partitioning the Sums of Squares
Just as in the case of simple linear regression, the sum of squares for the criterion
(UGPA in this example) can be partitioned into the sum of squares predicted and
the sum of squares error. That is,
SSY = SSY' + SSE
which for these data:
20.798 = 12.961 + 7.837
The sum of squares predicted is also referred to as the “sum of squares explained.”
Again, as in the case of simple regression,
Proportion Explained = SSY'/SSY
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In simple regression, the proportion of variance explained is equal to r2; in multiple
regression, the proportion of variance explained is equal to R2.
In multiple regression, it is often informative to partition the sums of squares
explained among the predictor variables. For example, the sum of squares
explained for these data is 12.96. How is this value divided between HSGPA and
SAT? One approach that, as will be seen, does not work is to predict UGPA in
separate simple regressions for HSGPA and SAT. As can be seen in Table 2, the
sum of squares in these separate simple regressions is 12.64 for HSGPA and 9.75
for SAT. If we add these two sums of squares we get 22.39, a value much larger
than the sum of squares explained of 12.96 in the multiple regression analysis. The
explanation is that HSGPA and SAT are highly correlated (r = .78) and therefore
much of the variance in UGPA is confounded between HSGPA or SAT. That is, it
could be explained by either HSGPA or SAT and is counted twice if the sums of
squares for HSGPA and SAT are simply added.
Table 2. Sums of Squares for Various Predictors
Predictors
Sum of Squares
HSGPA
12.64
SAT
9.75
HSGPA and SAT
12.96
Table 3 shows the partitioning of the sums of squares into the sum of squares
uniquely explained by each predictor variable, the sum of squares confounded
between the two predictor variables, and the sum of squares error. It is clear from
this table that most of the sum of squares explained is confounded between
HSGPA and SAT. Note that the sum of squares uniquely explained by a predictor
variable is analogous to the partial slope of the variable in that both involve the
relationship between the variable and the criterion with the other variable(s)
controlled.
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Table 3. Partitioning the Sum of Squares
Source
Sum of Squares
Proportion
HSGPA (unique)
3.21
0.15
SAT (unique)
0.32
0.02
HSGPA and SAT
(Confounded)
9.43
0.45
Error
7.84
0.38
Total
20.80
1.00
The sum of squares uniquely attributable to a variable is computed by comparing
two regression models: the complete model and a reduced model. The complete
model is the multiple regression with all the predictor variables included (HSGPA
and SAT in this example). A reduced model is a model that leaves out one of the
predictor variables. The sum of squares uniquely attributable to a variable is the
sum of squares for the complete model minus the sum of squares for the reduced
model in which the variable of interest is omitted. As shown in Table 2, the sum of
squares for the complete model (HSGPA and SAT) is 12.96. The sum of squares
for the reduced model in which HSGPA is omitted is simply the sum of squares
explained using SAT as the predictor variable and is 9.75. Therefore, the sum of
squares uniquely attributable to HSGPA is 12.96 - 9.75 = 3.21. Similarly, the sum
of squares uniquely attributable to SAT is 12.96 - 12.64 = 0.32. The confounded
sum of squares in this example is computed by subtracting the sum of squares
uniquely attributable to the predictor variables from the sum of squares for the
complete model: 12.96 - 3.21 - 0.32 = 9.43. The computation of the confounded
sums of squares in analyses with more than two predictors is more complex and
beyond the scope of this text.
Since the variance is simply the sum of squares divided by the degrees of
freedom, it is possible to refer to the proportion of variance explained in the same
way as the proportion of the sum of squares explained. It is slightly more common
to refer to the proportion of variance explained than the proportion of the sum of
squares explained and, therefore, that terminology will be adopted frequently here.
When variables are highly correlated, the variance explained uniquely by the
individual variables can be small even though the variance explained by the
variables taken together is large. For example, although the proportions of variance
502
explained uniquely by HSGPA and SAT are only 0.15 and 0.02 respectively,
together these two variables explain 0.62 of the variance. Therefore, you could
easily underestimate the importance of variables if only the variance explained
uniquely by each variable is considered. Consequently, it is often useful to consider
0.425
a set of related variables.
For example, assume you were interested in predicting
=
= 1.39of variables some of which reflect cognitive
job performance from a0.305
large number
ability. It is likely that these measures of cognitive ability would be highly
correlated among themselves and therefore no one of them would explain much of
the variance independent of the other variables. However, you could avoid this
2
problem by determining
=the proportion of variance explained by all of the
1
cognitive ability variables considered
together as a set. The variance explained by
the set would include all the variance explained uniquely by the variables in the set
as well as all the variance confounded among variables in the set. It would not
include variance confounded
0.627 5 with
2 variables outside the set. In short, you would be
=
computing the variance explained =
by 1.39
the set of variables that is independent of the
1 0.627
variables not in the set.
Inferential Statistics
1+
We begin by presenting
the formula for testing the significance of the contribution
= 0.5
1 show how special cases of this formula can be
of a set of variables. We will then
used to test the significance of R2 as well as to test the significance of the unique
contribution of individual variables.
1
The first step is to compute
two regression analyses: (1) an analysis in which
all the predictor variables are included
and (2) an analysis in which the variables in
3
the set of variables being tested are excluded. The former regression model is
called the “complete model” and the latter is called the “reduced model.” The basic
idea is
if the reduced
model explains much less than the complete model, then
Introduction
tothat
multiple
regression
the set of variables excluded from the reduced model is important.
The formula for testing the contribution of a group of variables is:
=
=
1
where:
503
SSQC is the sum of squares for the complete model,
SSQR is the sum of squares for the reduced model,
pC is the number of predictors in the complete
model,
pR is the number of predictors in the reduced
model,
SSQT is the sum of squares total (the sum of
squared deviations
of the criterion variable
SSQC − SSQ
R
from its mean), and
MSexplained
pC − pR
=
N is the totalSSQ
number
of observations
MSerror
T − SSQC
N − pC −1
F( pc− pr,N− pc−1) =
The degrees of freedom for the numerator is pc - pr and the degrees of freedom for
the denominator is N - pC -1. If the F is significant, then it can be concluded that
the variables excluded 12.96
in the reduced
− 9.75 set contribute to the prediction of the
criterion variable independently of the other
variables.
3.212
2
−1
F(1,102) =can be used to test the
= significance
= 41.80
This formula
of R2 by defining the
20.80
−12.96
0.077In this application, SSQR and pR =
reduced model as having no predictor variables.
105 − 2 −1
0. The formula is then simplified
as follows:
SSQC
pC SSQC MSexplained
F( pc,N−pc−1) =
=
SSQT − SSQp
MSerror MSexplained
CC
F( pc,N− pc−1) =
=
N −SSQ
pC −1
−
SSQ
MS
T
C
!
error
N − pC −1
which for this example becomes:
12.96
6.48
2
F=
=
= 84.35.
20.80 −12.96 0.08
105 − 2 −1
12.96 −12.64
0.322
2 −1
F=
=
= 4.19
20.80 −12.96 0.077
504
F( pc−pr,N−pc−1) =
C
R
SSQT − SSQC
N − pC −1
=
explained
MSerror
The degrees of freedom are 2 and 102. The F distribution calculator shows that p <
12.96 − 9.75
0.001.
3.212
2 −1to test the
The reduced
F(1,102) =model used
= variance= explained
41.80. uniquely by a single
20.80
−12.96
0.077
SSQ
predictor consists of all the
variables
except
the predictor variable in question. For
C − SSQ
R
example, the reduced model
a test of theMS
unique
contribution of HSGPA
105 −for
−1
p2
explained
C − pR
F
=
=
( pc−the
pr,N−
pc−1)
contains only
variable
SAT. Therefore,
theMS
sum of squares for the reduced
SSQ
T − SSQC
error
model is the sum of squares when UGPA is predicted by SAT. This sum of squares
N −SSQ
pC −1
C
is 9.75. The calculations for F are
shown
below:
MSexplained
pC
F( pc,N−pc−1)
=
=
12.96 SSQ
− 9.75
MSerror
T − SSQC
3.212
2 −1N − p= −1 = 41.80.
F(1,102) =
C
20.80 −12.96
105 − 2 −1
!
0.077
12.96
6.48
The degreesF=
of freedom2areSSQ
1 and
102.
The=F84.35.
distribution calculator shows that p <
C =
20.80 −12.96
0.001.
0.08MSexplained
pC
!
F
=
= test for the unique
( pc,N− pc−1)
Similarly,
the
reduced
model
in the
contribution of SAT
105
− 2 −1
SSQ
−
SSQ
MS
T
C
error
consists of HSGPA.
N − pC −1
12.96 −12.64
0.322
2 −1
F= 12.96
=
= 4.19.
6.48
20.80
−12.96
0.077
2
F=
=
= 84.35.
20.80
−12.96
0.08
105 − 2 −1
! 105 − 2 −1
The degrees! of freedom are 1 and 102. The F distribution calculator shows that p =
0.0432.
! 12.96 −12.64
The significance
the0.322
variance explained uniquely by a variable is
2 −1test of =
F=
= 4.19
identical to a significance
test of 0.077
the regression coefficient for that variable. A
20.80 −12.96
regression coefficient
105 −and
2 −1the variance explained uniquely by a variable both
reflect the! relationship between a variable and the criterion independent of the
other variables.
If the variance explained uniquely by a variable is not zero, then
!
the regression
coefficient cannot be zero. Clearly, a variable with a regression
!
coefficient of zero would explain no variance.
505
Other inferential statistics associated with multiple regression that are
beyond the scope of this text. Two of particular importance are (1) confidence
intervals on regression slopes and (2) confidence intervals on predictions for
specific observations. These inferential statistics can be computed by standard
statistical analysis packages such as R, SPSS, STATA, SAS, and JMP.
Assumptions
No assumptions are necessary for computing the regression coefficients or for
partitioning the sums of squares. However, there are several assumptions made
when interpreting inferential statistics. Moderate violations of Assumptions 1-3 do
not pose a serious problem for testing the significance of predictor variables.
However, even small violations of these assumptions pose problems for confidence
intervals on predictions for specific observations.
1. Residuals are normally distributed:
As in the case of simple linear regression, the residuals are the errors of
prediction. Specifically, they are the differences between the actual scores on
the criterion and the predicted scores. A Q-Q plot for the residuals for the
example data is shown below. This plot reveals that the actual data values at the
lower end of the distribution do not increase as much as would be expected for
a normal distribution. It also reveals that the highest value in the data is higher
than would be expected for the highest value in a sample of this size from a
normal distribution. Nonetheless, the distribution does not deviate greatly from
506
normality.
!0.5
Sample1Quantiles
0.0
0.5
Normal'Q)Q'Plot
!2
!1
0
1
Theoretical1Quantiles
2
2. Homoscedasticity:
It is assumed that the variance of the errors of prediction are the same for all
predicted values. As can be seen below, this assumption is violated in the
example data because the errors of prediction are much larger for observations
with low-to-medium predicted scores than for observations with high predicted
scores. Clearly, a confidence interval on a low predicted UGPA would
507
underestimate the uncertainty.
3. Linearity:
It is assumed that the relationship between each predictor variable and the
criterion variable is linear. If this assumption is not met, then the predictions
may systematically overestimate the actual values for one range of values on a
predictor variable and underestimate them for another.
508
Statistical Literacy
by David M. Lane
Prerequisites
• Chapter 14: Regression Toward the Mean
In a discussion about the Dallas Cowboy football team, there was a comment that
the quarterback threw far more interceptions in the first two games than is typical
(there were two interceptions per game). The author correctly pointed out that,
because of regression toward the mean, performance in the future is expected to
improve. However, the author defined regression toward the mean as, "In nerd
land, that basically means that things tend to even out over the long run."
What do you think?
Comment on that definition.
That definition is sort of correct, but it could be stated more
precisely. Things don't always tend to even out in the long run.
If a great player has an average game, then things wouldn't
even out (to the average of all players) but would regress toward
that player's high mean performance.
509
References
Schall, T., & Smith, G. (2000) Do Baseball Players Regress Toward the Mean? The
American Statistician, 54, 231-235.
510
Exercises
Prerequisites
All material presented in the Regression chapter
1. What is the equation for a regression line? What does each term in the line refer
to?
2. The formula for a regression equation is Y’ = 2X + 9.
a. What would be the predicted score for a person scoring 6 on X?
b. If someone’s predicted score was 14, what was this person’s score on X?
3. What criterion is used for deciding which regression line fits best?
4. What does the standard error of the estimate measure? What is the formula for
the standard error of the estimate?
5.
a. In a regression analysis, the sum of squares for the predicted scores is 100 and
the sum of squares error is 200, what is R2?
b. In a different regression analysis, 40% of the variance was explained. The sum
of squares total is 1000. What is the sum of squares of the predicted values?
6. For the X,Y data below, compute:
a. r and determine if it is significantly different from zero.
b. the slope of the regression line and test if it differs significantly from zero.
c. the 95% confidence interval for the slope.
511
7. What assumptions are needed to calculate the various inferential statistics of
linear regression?
8. The correlation between years of education and salary in a sample of 20 people
from a certain company is .4. Is this correlation statistically significant at the .
05 level?
9. A sample of X and Y scores is taken, and a regression line is used to predict Y
from X. If SSY’ = 300, SSE = 500, and N = 50, what is:
(a) SSY?
(b) the standard error of the estimate?
(c) R2?
10. Using linear regression, find the predicted post-test score for someone with a
score of 45 on the pre-test.
512
11. The equation for a regression line predicting the number of hours of TV
watched by children (Y) from the number of hours of TV watched by their
parents (X) is Y' = 4 + 1.2X. The sample size is 12.
513
a. If the standard error of b is .4, is the slope statistically significant at the .05
level?
b. If the mean of X is 8, what is the mean of Y?
12. Based on the table below, compute the regression line that predicts Y from X.
13. Does A or B have a larger standard error of the estimate?
14. True/false: If the slope of a simple linear regression line is statistically
significant, then the correlation will also always be significant.
15. True/false: If the slope of the relationship between X an Y is larger for
Population 1 than for Population 2, the correlation will necessarily be larger in
Population 1 than in Population 2. Why or why not?
16. True/false: If the correlation is .8, then 40% of the variance is explained.
17. True/false: If the actual Y score was 31, but the predicted score was 28, then
the error of prediction is 3.
Questions from Case Studies
Angry Moods (AM) case study
514
18. (AM) Find the regression line for predicting Anger-Out from Control-Out.
a. What is the slope?
b. What is the intercept?
c. Is the relationship at least approximately linear?
d. Test to see if the slope is significantly different from 0.
e. What is the standard error of the estimate?
SAT and GPA (SG) case study
19. (SG) Find the regression line for predicting the overall university GPA from the
high school GPA.
a. What is the slope?
b. What is the y-intercept?
c. If someone had a 2.2 GPA in high school, what is the best estimate of his or
her college GPA?
d. If someone had a 4.0 GPA in high school, what is the best estimate of his or
her college GPA?
Driving (D) case study
20. (D) What is the correlation between age and how often the person chooses to
drive in inclement weather? Is this correlation statistically significant at the .01
level? Are older people more or less likely to report that they drive in inclement
weather?
21. (D) What is the correlation between how often a person chooses to drive in
inclement weather and the percentage of accidents the person believes occur in
inclement weather? Is this correlation significantly different from 0?
22. (D) Use linear regression to predict how often someone rides public
transportation in inclement weather from what percentage of accidents that
person thinks occur in inclement weather. (Pubtran by Accident)
(a) Create a scatter plot of this data and add a regression line.
(b) What is the slope?
515
(c) What is the intercept?
(d) Is the relationship at least approximately linear?
(e) Test if the slope is significantly different from 0.
(f) Comment on possible assumption violations for the test of the slope.
(g) What is the standard error of the estimate?
516
``` |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Solving Exponential Equations
## Equations with terms raised to exponents including x
0%
Progress
Practice Solving Exponential Equations
Progress
0%
Solving Exponential Equations
"I'm thinking of a number," you tell your best friend. "The number I'm thinking of satisfies the equation 4x+1=256\begin{align*}4^{x + 1} = 256\end{align*}. What number are you thinking of?
### Guidance
Until now, we have only solved pretty basic exponential equations, like #1 in the Review Queue above. We know that x=5\begin{align*}x=5\end{align*}, because 25=32\begin{align*}2^5=32\end{align*}. Ones like #4 are a little more challenging, but if we put everything into a power of 2, we can set the exponents equal to each other and solve.
8x23x3xx=128=27=7=73
So, 873=128\begin{align*}8^{\frac{7}{3}} = 128\end{align*}.
But, what happens when the power is not easily found? We must use logarithms, followed by the Power Property to solve for the exponent.
#### Example A
Solve 6x=49\begin{align*}6^x=49\end{align*}. Round your answer to the nearest three decimal places.
Solution: To solve this exponential equation, let’s take the logarithm of both sides. The easiest logs to use are either ln\begin{align*}\ln\end{align*} (the natural log), or log (log, base 10). We will use the natural log.
6xln6xxln6x=49=ln49=ln49=ln49ln62.172
#### Example B
Solve 10x3=1003x+11\begin{align*}10^{x-3}=100^{3x+11}\end{align*}.
Solution: Change 100 into a power of 10.
10x3x3255=102(3x+11)=6x+22=5x=x
#### Example C
Solve 82x34=5\begin{align*}8^{2x-3}-4=5\end{align*}.
Solution: Add 4 to both sides and then take the log of both sides.
82x3482x3log82x3(2x3)log82x32xx=5=9=log9=log9=log9log8=3+log9log8=32+log92log82.56
Notice that we did not find the numeric value of log9\begin{align*}\log9\end{align*} or log8\begin{align*}\log8\end{align*} until the very end. This will ensure that we have the most accurate answer.
Intro Problem Revisit We can rewrite the equation 4x+1=256\begin{align*}4^{x + 1} = 256\end{align*} as 22(x+1)=28\begin{align*}2^{2(x+1)} = 2^8\end{align*} and solve for x.
22(x+1)=2822x+2=282x+2=8x=3
Therefore, you're thinking of the number 3.
### Guided Practice
Solve the following exponential equations.
1. 4x8=16\begin{align*}4^{x-8}=16\end{align*}
2. 2(7)3x+1=48\begin{align*}2(7)^{3x+1} =48\end{align*}
3. \begin{align*}\frac{2}{3} \cdot 5^{x+2}+9=21\end{align*}
1. Change 16 to \begin{align*}4^2\end{align*} and set the exponents equal to each other.
2. Divide both sides by 2 and then take the log of both sides.
3. Subtract 9 from both sides and multiply both sides by \begin{align*}\frac{3}{2}\end{align*}. Then, take the log of both sides.
### Explore More
Use logarithms and a calculator to solve the following equations for \begin{align*}x\end{align*}. Round answers to three decimal places.
1. \begin{align*}5^x = 65\end{align*}
2. \begin{align*}7^x = 75\end{align*}
3. \begin{align*}2^x = 90\end{align*}
4. \begin{align*}3^{x-2} = 43\end{align*}
5. \begin{align*}6^{x+1}+3=13\end{align*}
6. \begin{align*}6(11^{3x-2})=216\end{align*}
7. \begin{align*}8+13^{2x-5}=35\end{align*}
8. \begin{align*}\frac{1}{2} \cdot 7^{x-3}-5=14\end{align*}
Solve the following exponential equations without a calculator.
1. \begin{align*}4^x=8\end{align*}
2. \begin{align*}9^{x-2} = 27\end{align*}
3. \begin{align*}5^{2x+1}=125\end{align*}
4. \begin{align*}9^3=3^{4x-6}\end{align*}
5. \begin{align*}7(2^{x-3})=56\end{align*}
6. \begin{align*}16^x \cdot 4^{x+1}=32^{x+1}\end{align*}
7. \begin{align*}3^{3x+5}=3 \cdot 9^{x+3}\end{align*}
### Vocabulary Language: English
log
log
"log" is the shorthand term for 'the logarithm of', as in "$\log_b n$" means "the logarithm, base $b$, of $n$."
Logarithm
Logarithm
A logarithm is the inverse of an exponential function and is written $\log_b a=x$ such that $b^x=a$.
take the log of both sides
take the log of both sides
To take the log of both sides means to take the log of both the entire right hand side of the equation and the entire left hand side of the equation. As long as neither side is negative or equal to zero it maintains the equality of the two sides of the equation. |
# Equation of a Line Parallel to a Line
We will learn how to find the equation of a line parallel to a line.
Prove that the equation of a line parallel to a given line ax + by + λ = 0, where λ is a constant.
Let, ax + by + c = 0 (b ≠ 0) be the equation of the given straight line.
Now, convert the equation ax + by + c = 0 to its slope-intercept form.
ax + by+ c = 0
by = - ax - c
Dividing both sides by b, [b ≠ 0] we get,
y = -$$\frac{a}{b}$$ x - $$\frac{c}{b}$$, which is the slope-intercept form.
Now comparing the above equation to slope-intercept form (y = mx + b) we get,
The slope of the line ax + by + c = 0 is (- $$\frac{a}{b}$$).
Since the required line is parallel to the given line, the slope of the required line is also (- $$\frac{a}{b}$$).
Let k (an arbitrary constant) be the intercept of the required straight line. Then the equation of the straight line is
y = - $$\frac{a}{b}$$ x + k
by = - ax + bk
ax + by = λ, Where λ = bk = another arbitrary constant.
Note: (i) Assigning different values to λ in ax + by = λ we shall get different straight lines each of which is parallel to the line ax + by + c = 0. Thus, we can have a family of straight lines parallel to a given line.
(ii) To write a line parallel to a given line we keep the expression containing x and y same and simply replace the given constant by a new constant λ. The value of λ can be determined by some given condition.
To get it more clear let us compare the equation ax + by = λ with equation ax + by + c = 0. It follows that to write the equation of a line parallel to a given straight line we simply need to replace the given constant by an arbitrary constant, the terms with x and y remain unaltered. For example, the equation of a straight line parallel to the straight line 7x - 5y + 9 = 0 is 7x - 5y + λ = 0 where λ is an arbitrary constant.
### Solved examples to find the equations of straight lines parallel to a given line:
1. Find the equation of the straight line which is parallel to 5x - 7y = 0 and passing through the point (2, - 3).
Solution:
The equation of any straight line parallel to the line 5x - 7y = 0 is 5x - 7y + λ = 0 …………… (i) [Where λ is an arbitrary constant].
If the line (i) passes through the point (2, - 3) then we shall have,
5 ∙ 2 - 7 ∙ (-3) + λ = 0
10 + 21 + λ = 0
31 + λ = 0
λ = -31
Therefore, the equation of the required straight line is 5x - 7y - 31 = 0.
2. Find the equation of the straight line passing through the point (5, - 6) and parallel to the straight line 3x - 2y + 10 = 0.
Solution:
The equation of any straight line parallel to the line 3x - 2y + 10 = 0 is 3x - 2y + k = 0 …………… (i) [Where k is an arbitrary constant].
According to the problem, the line (i) passes through the point (5, - 6) then we shall have,
3 ∙ 5 - 2 ∙ (-6) + k = 0
15 + 21 + k = 0
36 + k = 0
k = -36
Therefore, the equation of the required straight line is 3x - 2y - 36 = 0. |
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# Evaluate: $\int {\dfrac{{1 + \log x}}{{x\left( {2 + \log x} \right)\left( {3 + \log x} \right)}}} {\text{dx}}$
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Hint- First use substitution method and then partial fraction method to simplify the integral.
Here we have to evaluate $\int {\dfrac{{1 + \log x}}{{x\left( {2 + \log x} \right)\left( {3 + \log x} \right)}}} {\text{dx}}$
So let’s substitute ${\text{1 + logx = p}}$
So on differentiating both the sides we have $\left( {{\text{0 + }}\dfrac{1}{x}} \right)dx = dp$
Let’s make this substitution back into our main integral we get
$\int {\dfrac{p}{{x\left( {1 + p} \right)\left( {2 + p} \right)}} \times xdp}$ As ${\text{(2 + logx)}}$can be written as ${\text{(1 + (1 + logx))}}$and $\left( {3 + \log x} \right)$can be written as $\left( {2 + (1 + \log x)} \right)$
On simplifying we get
$\int {\dfrac{p}{{\left( {1 + p} \right)\left( {2 + p} \right)}}} {\text{dp}}$
Now let’s resolve it into partial fractions so we can write this form as
Let $\dfrac{p}{{\left( {1 + p} \right)\left( {2 + p} \right)}} = \dfrac{A}{{1 + p}} + \dfrac{B}{{2 + p}}$……………………………. (1)
So let’s take LCM in the right side we get
$\dfrac{p}{{\left( {1 + p} \right)\left( {2 + p} \right)}} = \dfrac{{A\left( {2 + p} \right) + B\left( {1 + p} \right)}}{{\left( {1 + p} \right)\left( {2 + p} \right)}}$
Denominator in both sides will cancel it out so we get
${\text{p = }}A\left( {2 + p} \right) + B\left( {1 + p} \right)$……………………………………. (2)
Now let’s put ${\text{p = - 2}}$ so that we can find the value of B
We get
${\text{ - 2 = 0 - B}}$ Hence our ${\text{B = 2}}$………………………………………. (3)
Now we need to find A so let’s put ${\text{p = - 1}}$ in equation (2) we get
${\text{ - 1 = A + 0}}$ Hence our ${\text{A = - 1}}$………………………………… (4)
Now let’s put equation (3) and equation (4) back into (1) we get
$\dfrac{p}{{\left( {1 + p} \right)\left( {2 + p} \right)}} = \dfrac{{ - 1}}{{1 + p}} + \dfrac{2}{{2 + p}}$
So our integral can now be written as
${\text{I}} = \int {\left( {\dfrac{{ - 1}}{{1 + p}} + \dfrac{2}{{2 + p}}} \right){\text{ }}} {\text{dp}}$
We can segregate this integral as
${\text{I = }}\int {\dfrac{{ - 1}}{{1 + p}}{\text{ dp}}} {\text{ + }}\int {\dfrac{2}{{2 + p}}} {\text{ dp}}$
Now using the standard formula for integral of$\int {\dfrac{{dx}}{x} = \log \left| x \right|}$, we can solve above as
${\text{I = - log}}\left| {1 + p} \right|{\text{ + 2log}}\left| {2 + p} \right|{\text{ + c}}$
Let’s substitute back the value of p which was ${\text{1 + logx}}$ we get
${\text{I = - log}}\left| {2 + \log x} \right| + 2\log \left| {3 + \log x} \right|{\text{ + c}}$
Using the property of log that is ${\text{log(A) - log(B) = log}}\left( {\dfrac{A}{B}} \right)$and ${\text{nlog(x) = log(x}}{{\text{)}}^n}$
${\text{I = log}}\left| {\dfrac{{{{\left( {\log x + 3} \right)}^2}}}{{\log x + 2}}} \right|{\text{ + c}}$
Note- All such type of problems are based upon the concept of evaluation by substitution, whenever we see some terms in the integral that are related or can be converted into one another in numerator or in denominator then we can use the basic term and substitute it to some quantity in order to simplify the integral.
Last updated date: 01st Jun 2023
Total views: 330.3k
Views today: 6.86k |
# Min-Coin Change
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
The Minimum Coin Change (or Min-Coin Change) is the problem of using the minimum number of coins to make change for a particular amount of cents, $n$, using a given set of denominations $d_{1}\ldots d_{m}$. This is closely related to the Coin Change problem.
## Overview
The problem is typically asked as: If we want to make change for $N$ cents, and we have infinite supply of each of $S=\{S_{1},S_{2},\ldots ,S_{m}\}$ valued coins, what is the least amount of coins we need to make the change? (For simplicity's sake, the order does not matter.)
Mathematically, we are trying to minimize $\sum {x_{k}}$ for $N=\sum _{{k=1\ldots m}}{x_{k}S_{k}}$ where $x_{k}\geq 0,k\in \{1\ldots m\}$
## Greedy Approach
There are special cases where the greedy algorithm is optimal - for example, the US coin system. However this is not true in the general case. An example of a counterexample is:
Given the denominations 1, 5, 10, 20, 25, and wish to make change for 40 cents, the greedy algorithm would give us 25, 10, 5, but the best solution only requires 2 coins - 2 of the 20 cent coins.
## Recursive Formulation
$C(N,m)=\min {(C(N,m-1),C(N-S_{m},m)+1)}$
with the base cases:
• $C(N,m)=0,N=0$
• $C(N,m)=\infty ,N<0$
• $C(N,m)=\infty ,N\geq 1,m\leq 0$
If the result of $C(N,m)$ is either $0$ or $\infty$ then it is impossible make change for $N$ with the given coins. |
# Weights and Measures
By | 2011-03-20
You have a set of balance scales and some weights. You need to be able to measure every whole number weight from 1 to 40 kilograms of (say) flour. What is the smallest number of weights you need and what are their values?
The first trick is of course that you can put weights on either side, so you have to notice that you don’t necessarily need to be able to sum to every value, you can subtract as well. If we put the flour on the right hand side, then anything we put on the left adds to the balance, and anything we put on the right subtracts from the balance. For example, if we had two weights, 1kg and 3kg we could make.
• 1kg
• 3kg
• 1kg + 3kg = 4kg
• 1kg – 3kg = –2kg
• 3kg – 1kg = 2kg
Obviously the negative weight isn’t going to be useful to us, as we will never be called upon to weigh –2kg of flour.
So, each weight can be in one of three places:
• The same side as the flour
• Not the same side as the flour
• Not on the scale at all
We’ll call the side the flour is on the “minus-side” (since, from the point of view of the plus side, it effectively lowers the weight of the plus side) and the other side the “plus-side”.
$Y$
and
$Z$
, here are all possible configurations:
• $0$
• $+Y–0$
• $+Z–0$
• $0–Y$
• $0–Z$
• $Y–Z$
• $Z–Y$
• $+Y+Z–0$
• $0–Y–Z$
There are nine configurations, which shouldn’t surprise us. Two weights each in any of three places, is
$3×3=9$
Let’s also recall that we can’t measure 0kg and we can’t measure negative weights of flour (since they cannot exist), therefore our equation for real combinations measurable with
$m$
weights is
$N\left(m\right)=\frac{\left({3}^{m}–1\right)}{2}$
We can therefore be certain then that two weights isn’t sufficient to measure 40 different weights of flour. Similarly, three weights would be
$3×3×3=27$
$N\left(3\right)=\frac{\left(27–1\right)}{2}=13$
Thirteen weights would be insufficient to distinguish between 40 different inputs.
$3×3×3×3=81$
$N\left(4\right)=\frac{\left(81–1\right)}{2}=40$
Perfect! What we want is possible. Let’s list all the configurations of four weights (zero and negatives included). In the following table, “0” means not on the scales, “-” means on the same side as the flour and “+” means not on the same side as the flour (the eagle eyed amongst you will notice that this is simply a base–3 count):
``````W X Y Z W X Y Z W X Y Z
------------- ------------- -------------
+ + + + - + + + 0 + + +
+ + + - - + + - 0 + + -
+ + + 0 - + + 0 0 + + 0
+ + - + - + - + 0 + - +
+ + - - - + - - 0 + - -
+ + - 0 - + - 0 0 + - 0
+ + 0 + - + 0 + 0 + 0 +
+ + 0 - - + 0 - 0 + 0 -
+ + 0 0 - + 0 0 0 + 0 0
+ - + + - - + + 0 - + +
+ - + - - - + - 0 - + -
+ - + 0 - - + 0 0 - + 0
+ - - + - - - + 0 - - +
+ - - - - - - - 0 - - -
+ - - 0 - - - 0 0 - - 0
+ - 0 + - - 0 + 0 - 0 +
+ - 0 - - - 0 - 0 - 0 -
+ - 0 0 - - 0 0 0 - 0 0
+ 0 + + - 0 + + 0 0 + +
+ 0 + - - 0 + - 0 0 + -
+ 0 + 0 - 0 + 0 0 0 + 0
+ 0 - + - 0 - + 0 0 - +
+ 0 - - - 0 - - 0 0 - -
+ 0 - 0 - 0 - 0 0 0 - 0
+ 0 0 + - 0 0 + 0 0 0 +
+ 0 0 - - 0 0 - 0 0 0 -
+ 0 0 0 - 0 0 0 0 0 0 0
``````
Let us (for our own convenience) assume that
$W$
is the heaviest, running to
$Z$
, the lightest. Such that
$W\ge X\ge Y\ge Z$
We know that the most we can put on one side is all the weights. Therefore all of our weights must add up to some maximum,
$N$
.
$W+X+Y+Z=N$
The second heaviest weight we are asked to weigh is
$N–1$
, and the least we can take off is the smallest weight,
$Z$
, therefore
$W+X+Y=N–1$
Therefore
$Z=1$
(subtract these two equations). Which then enables us to measure
$N–2$
by putting
$Z$
on the minus-side.
Now we must measure
$N–3$
. The least we can do is replace
$Z$
on the plus-side and remove
$Y$
.
$W+X+Z=N–3$
Therefore
$Y=3$
. Which then enables us to measure
$\left(N–4\right)$
,
$\left(N–5\right)$
,
$\left(N–6\right)$
,
$\left(N–7\right)$
, and
$\left(N–8\right)$
. Recapping:
$W+X+Y+Z=N$
$W+X+Y+0=N–1$
$W+X+Y–Z=N–2$
$W+X+0+Z=N–3$
$W+X+0+0=N–4$
$W+X+0–Z=N–5$
$W+X–Y+Z=N–6$
$W+X–Y+0=N–7$
$W+X–Y–Z=N–8$
You should be able to see the pattern from the table evolving now. Each weight is either positive, off, or negative. Which means we get double duty from each one. Taking
$Z$
off lowers the total by 1kg; adding
$Z$
to the negative side lowers the total by another 1kg. So, with
$Z$
we can subtract 0, 1 or 2 from a given point (where
$Z$
is “on”).
$Y$
lets us subtract 0, 3 or 6 from a given point (where
$Y$
is “on”). Combined
$Z$
and
$Y$
lets us subtract 0, 1, 2, 3, 4, 5, 6, 7, or 8 from a given point (where
$Y$
and
$Z$
are “on”).
It’s now pretty simple to carry on. The next smallest change we can make is to remove the second heaviest weight; we want that to give us
$\left(N–9\right)$
:
$W+0+Y+Z=N–9$
Therefore
$X=9$
.
$Y$
and
$Z$
are both “on” so we know that we can subtract another 8 kg from
$\left(N–9\right)$
in sequence, taking us to:
$W+0+Y+0=N–10$
$W+0+Y–Z=N–11$
$W+0+0+Z=N–12$
$W+0+0+0=N–13$
$W+0+0–Z=N–14$
$W+0–Y+Z=N–15$
$W+0–Y+0=N–16$
$W+0–Y–Z=N–17$
Now we have to put X on the minus side, from there we know that
$Y$
and
$Z$
back on will give us the ability to do another 8 kg:
$W–X+Y+Z=N–18$
$\dots$
$W–X–Y–Z=N–26$
That only leaves us with the option of putting
$X$
,
$Y$
and
$Z$
back on, and taking
$W$
off, which we want to give us the next value,
$N–27$
$0+X+Y+Z=N–27$
Subtracting this from the initial equation,
$\left(W+X+Y+Z=N\right)$
:
$W=N–\left(N–27\right)$
$=27$
So,
$W=27$
. Further,
$W$
,
$X$
,
$Y$
and
$Z$
are now known, so we can calculate that
$N=40$
.
We already know that we can position
$X$
,
$Y$
and
$Z$
to subtract up to 26, which is just what we need to get us to 1kg.
$N=40$
$W=27$
$X=9$
$Y=3$
$Z=1$
QED.
Now let’s go even more general. For
$n$
weights, what is the maximum weight of flour we can weigh? What weights do we need?
We’ll plough through this a little faster, first let’s add an additional weight:
$V+W+X+Y+Z=N$
$W$
,
$X$
,
$Y$
and
$Z$
are, their derivation is utterly unchanged by the presence of
$V$
. We run out when we come to:
$V–W–X–Y–Z=N–80$
I know that it’s
$N–80$
as follows:
$V+W+X+Y+Z=N$
(eq1)
$W+X+Y+Z=40$
(eq2)
$V–W–X–Y–Z=N–80$
(eq1)-
$2×$
(eq2)
As before, we put all
$W$
,
$X$
,
$Y$
and
$Z$
back on the plus side, and take
$V$
off, and that must be equal to our next integer weight:
$0+W+X+Y+Z=N–81$
As before, we subtract this from our initial equation,
$V+W+X+Y+Z=N$
:
$V+W–W+X–X+Y–Y+Z–Z=N–\left(N–81\right)$
$V=81$
We shouldn’t be surprised; as we said, each weight can be used twice, once when we take it off the plus side, and once when we add it to the minus side. Therefore, the nth weight is the sum of all the previous weights multiplied by two, plus one to get us to the next unknown.
Now we’ll label the
$n$
th weight
$W\left(n\right)$
, with
$n=0$
being the first weight, and
$W\left(0\right)$
equal to one.
$W\left(n\right)$
can be expressed as a function of the sum of all the previous weights:
$W\left(n\right)=\left(\sum _{i=0}^{n–1}W\left(i\right)\right)×2+1$
So:
$W\left(0\right)=1$
$W\left(1\right)=W\left(0\right)×2+1=3$
$W\left(2\right)=\left(W\left(0\right)+W\left(1\right)\right)×2+1=9$
$W\left(3\right)=\left(W\left(0\right)+W\left(1\right)+W\left(2\right)\right)×2+1=27$
$W\left(4\right)=\left(W\left(0\right)+W\left(1\right)+W\left(2\right)+W\left(3\right)\right)×2+1=81$
We should note that the maximum we can weigh with
$n$
weights is the sum of those
$n$
weights, or
$N\left(n\right)=\sum _{i=0}^{n}W\left(i\right)$
Therefore,
$W\left(n\right)=N\left(n–1\right)×2+1$
We should also note that we can get rid of the summation, and simply define
$N\left(n\right)$
in terms of
$N\left(n–1\right)$
:
$N\left(n\right)=N\left(n–1\right)+W\left(n\right)$
Substituting for
$W\left(n\right)$
:
$N\left(n\right)=N\left(n–1\right)+\left(N\left(n–1\right)×2+1\right)$
$=3N\left(n–1\right)+1$
$N\left(0\right)$
we have to define as 1 to match our
$W\left(0\right)$
defined as 1.
$N\left(n\right)=1,4,13,40,121,364,\dots$
$W\left(n\right)=1,3,9,27,81,243,729,\dots$
We’re still defining each term in terms of the previous term, which we needn’t do. We’ve already got an equation for
$N\left(n\right)$
:
$N\left(n\right)=\frac{\left({3}^{n+1}–1\right)}{2}$
This is the very first equation I gave, but adjusted for a start-from-zero
$n$
$m$
. Substituting this into our equation for
$W$
:
$W\left(n\right)=\left(\frac{{3}^{n–1+1}–1}{2}\right)×2+1$
$={3}^{n}$
This shouldn’t be a surprise to us really. The weights are a base–3 counting system, for any base, the nth digit’s multiplier is given by
${B}^{n}$
. If we’d been really clever, we would have recognised this right from the start and just gone straight to the answer. Oh well. |
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# JEE - Differential Calculus
Last updated date: 19th Sep 2024
Total views: 106.2k
Views today: 3.06k
## Introduction
Differential calculus is concerned with the rate at which one quantity changes in relation to another. You can also think of it as a study of quantity change rates. For instance, velocity is the rate at which a distance changes with respect to time in a specific direction. If f(x) is a function, the differential equation is f'(x) = dy/dx, where f'(x) is the function's derivative, y is the dependent variable, and x is the independent variable.
### Important Terms in Differential Calculus
• Function
• Variable
• Domain and Range
• Limits
• Continuity
• Differentiability
• Differentiation formulae
• Tangent and Normal
• Approximation
• Rate measure
• Maxima and Minima
• Differential Equation
### Limit
In calculus, the limit is extremely significant. In calculus, limits are used to define continuity, integrals, and derivatives. A function's limit is defined as follows:
Let's say the function "f" is defined on an open interval containing some numbers, such as "a," with the exception of "a" itself. The limit of a function f(x) is written as: $\begin{array}{l}\lim_{x\rightarrow a}f(x)= L\end{array}$ ,iff given e > 0, there exists d > 0 such that 0 < |x – a| < d implies that |f(x) – L| < e It means that the limit f(x) as “x” approaches “a” is “L”.
### Continuity
Continuity of a function states the characteristics of the function and its functional value. A function is said to be continuous if the curve has no missing points or breaking points in a given interval or domain, that is the curve is continuous at every point in its domain.
A function f(x) is known as a continuous function at a point x = a, in its domain if the following listed three conditions are satisfied-
1. f (a) exists which means that the value of f (a) is finite.
2. $\displaystyle \lim_{x \to a} f(x)$ exists, that is the right-hand limit = left-hand limit, and both R.H.S and L.H.S are finite.
3. $\displaystyle \lim_{x \to a} f(x)$ = f (a)
A function f(x) is said to be continuous in the given interval I that is equal to [x1, x2] only if the three conditions listed above are satisfied for every point in the given interval I.
### Formal Definition of Continuity
A function is said to be continuous in the closed interval (a,b) if:
1. f is continuous in (a, b)
2. $\displaystyle \lim_{x \to a^{+}} f(x) = f(a)$
3. $\displaystyle \lim_{x \to a^{1}} f(x)= f(a)$
A function is said to be continuous in the open interval (a, b) if, f (x) is going to be continuous within the unbounded interval (a, b) if at any point within the given interval the function is continuous.
### Differentiability
Function f(x) is said to be differentiable at the point x = a and if the derivative of the function f ‘(a) exists at every point in its given domain.
Differentiability Formula
The differentiability formula is defined by -
f’(a) = $\frac{f(a+h)-f(a)}{h}$
If a function is continuous at a particular point then a function is said to be differentiable at any point x = a in its domain. The vice versa of this is not always true.
### Derivative
1.
$\frac{d}{dx}\left ( sin \, x \right )$
$cos \, x$
2.
$\frac{d}{dx}\left ( cos \, x \right )$
$- sin \, x$
3.
$\frac{d}{dx}\left ( tan \, x \right )$
$sec^{2}x$
4.
$\frac{d}{dx}\left ( cot \, x \right )$
$- cosec^{2}x$
5.
$\frac{d}{dx}\left ( sec \, x \right )$
$sec \, x.\, tan \, x$
6.
$\frac{d}{dx}\left ( cosec \, x \right )$
$- cosec \, x.\, cot \, x$
7.
$\frac{d}{dx}\left ( sin^{-1}x \right )$
$\frac{1}{\sqrt{1-x^{2}}}$
8.
$\frac{d}{dx}\left ( cos^{-1}x \right )$
$-\frac{1}{\sqrt{1-x^{2}}}$
9.
$\frac{d}{dx}\left ( tan^{-1}x \right )$
$\frac{1}{1+x^{2}}$
10.
$\frac{d}{dx}\left ( cot^{-1}x \right )$
$- \frac{1}{1+x^{2}}$
11.
$\frac{d}{dx}\left ( sec^{-1}x \right )$
$\frac{1}{\left|x \right|\sqrt{x^{2}-1}}$
12.
$\frac{d}{dx}\left ( cosec^{-1}x \right )$
$-\frac{1}{\left|x \right|\sqrt{x^{2}-1}}$
13.
$\frac{d}{dx}\left ( a^{x} \right )$
$a^{x}\, lnx$
14.
$\frac{d}{dx}\left ( e^{x} \right )$
$e^{x}$
15.
$\frac{d}{dx}\left ( ln\, x \right )$
$\frac{1}{x}$
### Tangent and Normal
If tangent at P makes an angle $\theta$ with positive direction of x - axis, then slope of the tangent is $m = tan\theta$. Also, the slope of the tangent to the curve at any point on it is $\frac{dy}{dx}$.
$\therefore m = tan\theta =\frac{dy}{dx}$.
Also,the slope of normal is $-\frac{dx}{dy}$.
Therefore, equation of tangent at point $P\left ( x_{1}, \,\ y_{1} \right )$
$\Rightarrow \left ( y-y_{1} \right )= \frac{dy}{dx}_{\left ( x_{1}, \,\ y_{1} \right )}\left ( x_{1}, \,\ y_{1} \right )$.
And the equation of the normal at point $P\left ( x_{1}, \,\ y_{1} \right )$
$\Rightarrow \left ( y-y_{1} \right )= -\frac{dx}{dy}_{\left ( x_{1}, \,\ y_{1} \right )}\left ( x_{1}, \,\ y_{1} \right )$.
### Approximation Using Derivative
Consider the following figure to recall the definition of derivatives of function $y = f(x)$ at some point $P\left ( x, \,\ f(x) \right )\,\ w.r.to..x$.
From the definition of derivatives, we have
$\frac{dy}{dx}=\displaystyle \lim_{x \to 0}\frac{\Delta y}{\Delta x}= \displaystyle \lim_{x \to 0}\frac{f\left ( x+\Delta x \right )-f(x)}{\Delta x}$.
Here, when point Q moves closer to point P on the curve, then though $\Delta x$ does not tend to zero, $\Delta y$.
In that case,
$\Rightarrow \frac{dy}{dx}=\frac{\Delta y}{\Delta x}= \frac{f\left ( x+\Delta x \right )-f(x)}{\Delta x}$.
$\therefore f\left ( x+\Delta x \right )=f(x)+\Delta x \,\ \frac{dy}{dx}$.
### Monotonicity
One of the most important tools in analyzing the functions is their monotonic behavior. Functions like $f(x)= 2x+3, \,\ g(x)= log_{e}x \,\ and \,\ h(x) = x^{3}$ increase with the increase in the input value of the independent variable in their entire domain, i,e., graph of the function rises from left to right. Functions like $f(x)= cos^{-1}x$ and $g(x)= 0.5^{x}$ always decrease with increase in the value of x. Both of the above categories of functions are called monotonic functions.
Classification of functions based on monotonicity
• Increasing Function
• Strictly Increasing Function
• Decreasing Function
• Strictly Decreasing Function
### Maxima and Minima
Consider the graph of one differential functions as shown in the following figure
We observe that points A, B and C are stationary points where tangent to the curve is parallel to the x axis i.e., the derivative is zero. These two stationary points are referred to as turning points.
Point A in figure is called a point of local maxima because in its immediate area it is the highest point, and so represents the greatest or maximum value of the function in that area.
Point B is called a local minimum because in its immediate area it is the lowest point, and so represents the least or minimum value of the function in that area.
### Solved Examples:
Example 1: If the equation of tangent to the curve $y^{2}=ax^{3}+b$ at point (2, 3) is y = 4x - 5, then find the values of a and b
Solution:
Here, we have $y^{2}=ax^{3}+b$ ……………(1)
Differentiating w.r.t, x , we get
$\Rightarrow 2y\frac{dy}{dx}= 3ax^{2}$
$\Rightarrow \frac{dy}{dx}=\frac{3ax^{2}}{2y}$
$\left ( \frac{dy}{dx} \right )_{\left ( 2,3 \right )}=\frac{3\times a\times \left ( 2 \right )^{2}}{2\times 3}=2a$
Now, line y = 4x - 5 is tangent to the curve at (2, 3).
$\therefore 2a = 4$
$\Rightarrow a = 2$
Also, (2, 3) lies on the curve(1)
$\Rightarrow 9 = 8a+b$
$\therefore b=-7$
Example 2: If the graph of the function $f(x) = 3x^{4}+2x^{3}+ax^{2}-x+2$ is concave upward for all real x, then find the values of a.
Solution:
Here, we have $f'(x) = 12x^{3}+6x^{2}+2ax-1$
And $f''(x) = 36x^{2}+12x+2a$
Now graph of concave upward for all real x.
$\therefore f''(x)=36x^{2}+12x+2a>0$ for all real x.
Then, discriminant, $12^{2}-4(36)(2a)<0$
$\Rightarrow 1-2a<0$
$\Rightarrow a >\frac{1}{2}$
### Solved Problems of Previous Year Question
1. Let f, $f,g:\left [ -1,2 \right ]\to R$ be continuous functions which are twice differentiable on the interval (-1, 2). Let the values of f and g at the points -1, 0 and 2 be as given in the following table:
x = -1 x=0 x=2 f(x) 3 6 0 g(x) 0 1 -1
In each of the intervals (-1, 0) and (0, 2), the function (f-3g)’’ never vanishes. Then the correct statements is are
1. $f'(x)-3g'(x)=0$ has exactly three solutions in $\left ( -1,0 \right )\cup \left ( 0,2 \right )$.
2. $f'(x)-3g'(x)=0$ has exactly one solution in (-1, 0)
3. $f'(x)-3g'(x)=0$ has exactly one solution in (0, 2)
4. $f'(x)-3g'(x)=0$ has exactly two solutions in (-1, 0) and exactly two solutions in (0, 2)
Ans: (b), (c)
Let $h(x)=f(x)-3g(x)$
$h(-1)= f(-1)-3g(-1)=3-0=3$
$h(0)= f(0)-3g(0)=6-3=3$
$h(2)= f(2)-3g(2)=0-(-3)=3$
Thus, $h'(x)= 0$ has at least one root in (-1 , 1) and at least one root in (0, 2).
But Since $h''(x)= 0$ has no root in (-1, 0) and (0, 2) therefore $h'(x)= 0$ has exactly 1 root in (-1, 0) and exactly 1 root in (0, 2).
2. The function $f(x)=2\left|x \right|+\left|x +2 \right|+\left|\left|x+2 \right|-2\left|x \right| \right|$ has a local minimum or a local maximum at x =
1. $-2$
2. $\frac{-2}{3}$
3. $2$
4. $\frac{2}{3}$
$\Rightarrow \frac{f(x)=2\left|x \right|+\left|x +2 \right|+\left|\left|x+2 \right|-2\left|x \right| \right|}{2}=Min \, \left ( \left| 2x\right|, \left|x+2 \right| \right )$.
From the figure shown, points of local minima/ maxima are x = - 2, -⅔. 0
3. Let $a\in R$ and let $f:R\to R$ be given by $f(x)=x^{5}-5x+a$, then
1. f(x) has three real roots if a > 4
2. f(x) has only one real root if a > 4
3. f(x) has three real roots if a < - 4
4. f(x) has three roots if - 4 < a < 4
Ans: b, d
Let $y= f(x)=x^{5}-5x$
$\Rightarrow f'(x)= 5x^{4}-5$
$= 5(x-1)(x+1)(x^{2}+1)$
$f'(x)=0, \, \therefore x=-1,1$
$f''(x)= 20x^{3}$
$f''(1)= 20$ and $f''(-1)= -20$
Graph of y = f(x) is as shown in the following figure
From the graph $x^{5}-5x=-a$ has one real root if $-a<-4$ or $-a>4$
$\Rightarrow a>4 \,\ or \,\ a<-4$
$x^{5}-5x = -a$ has three real roots if -4 < -a < 4 i.e., -4 < a < 4
### Practice Question:
1. If f : R is a differentiable function such that $f'(x)> 2f(x)$ for all $x\in R$, and f(0) = 1, then
1. $f(x)>e^{2x}$ in $\left ( 0, \infty \right )$
2. f(x) is decreasing in $\left ( 0, \infty \right )$
3. f(x) is increasing in $\left ( 0, \infty \right )$
4. $f'(x)<e^{2x}$ in $\left ( 0, \infty \right )$
2. The function $f : \left [ 0, 3 \right ]\to \left [ 1, 29 \right ]$ defined by $f(x)= 2x^{3}-15x^{2}+36x+1$, is
1. One - one and onto
2. Onto but not one one
3. One - one but not onto
4. Neither one - one nor onto
### Conclusion:
In this chapter, we have elaborated on concepts and solutions to questions on the topic of Differential Calculus. Everything you're looking for is available in a single location. Students can carefully read through the Concepts, Definitions, and questions in the PDFs, which are also free to download and understand the concepts used to solve these questions. This will be extremely beneficial to the students in their exams.
## FAQs on JEE - Differential Calculus
1. What are some applications of Derivatives in real-life examples?
Almost all the applications have some real-life usage when it comes to partial derivatives and absolute derivatives. When a value y varies with x such that it satisfies y=f(x), then f’(x) = dy/dx is called the rate of change of y with respect to x. Also, f’(x0) = dy/dx x=x0 is the rate of change of y with respect to x=x0. The rate of change of values is a significant application of differentiation, which is used broadly in physics and other engineering subjects.
2. What is the definition of tangent in Mathematics?
The tangent is a straight line that just touches the curve at a given point. The normal is a straight line that is perpendicular to the tangent. The tangent to any curve at a given point is observed to have the same gradient as the curve at that point. A derivative is required in order to find the gradient of the curve, which can be found by differentiating and obtaining an expression for dy/dx. The gradient of the tangent can be achieved by substituting an x-value. To find the full equation of the tangent, it remains to find the y-intercept of the tangent, which can be found using the coordinates of the point that the tangent and the curve have in the common. |
# Square & Square Root of 1444
Created by: Team Maths - Examples.com, Last Updated: April 30, 2024
## Square & Square Root of 1444
In mathematics, particularly in algebra, the concepts of square and square roots are fundamental. Squaring a number, like 1444, means multiplying it by itself, yielding a result such as 2,085,136. This operation is crucial in exploring properties of both rational numbers (which can be expressed as a fraction of two integers) and Irrational Numbers (which cannot be neatly expressed as a fraction). Understanding these concepts helps deepen knowledge of mathematical relationships and patterns.
## Square of 1444
1444² (1444×1444) = 2085136
This calculation is what is referred to as “squaring” the number 1444, and the result, 2,085,136, is the “square” of 1444. This process is used in various mathematical applications to represent area or to perform computations in algebra and geometry.
## Square Root of 1444
√1444 = 38
The square root of a number is a value that, when multiplied by itself, gives the original number. In this case, the number is 1444.
Finding the square root of 1444 involves determining what number, when multiplied by itself, equals 1444. Mathematically, this is represented as:
1444 = 38
This means that 38 multiplied by 38 equals 1444:
38×38 = 1444
The square root operation is widely used in mathematics, particularly in algebra and geometry, to solve various equations and problems involving areas and lengths.
Square Root of 1444: 38
Exponential Form: 1444^½ or 1444^0.5
## Is the Square Root of 1444 Rational or Irrational?
The square root of 1444 is an Rational Number.
The square root of 1444 is rational. A number is considered rational if it can be expressed as the fraction of two integers (where the denominator is not zero). In this case, the square root of 1444 is 38, which can be expressed as a fraction 38/1. Since 38 is an integer, its square root is a rational number.
## Methods to Find Value of Root 1444
Finding the square root of 1444 can be approached using several methods, each with its own application and level of precision. Here are some common methods:
Prime Factorization:
Break down 1444 into its prime factors. Since 1444 = 2 × 2 × 19 × 19, its square root is the product of each factor taken once, i.e., √1444 = 2×19 =3 8.
Long Division Method:
This is a manual method that involves a step-by-step division-like process to find the square root. It’s particularly useful for larger numbers or when the prime factors aren’t easily apparent.
Using a Calculator:
The most straightforward method is to use a calculator, either a physical one or a software-based one like those on computers and smartphones.
Newton’s Method (Newton-Raphson):
This is a numerical technique for finding approximate solutions to numerically solving equations like 𝑥²−1444 = 0. It involves iterative guessing and refining the guess based on the derivative of the function.
Babylonian Method (Guess and Check):
This ancient method, also known as the “Heron’s method”, involves making an initial guess at the square root and then improving the guess and Repeat until the value stabilizes.
## Square Root of 1444 by Long Division Method
Pairing the Number:
• Divide the number 1444 into pairs from right to left. In this case, you have two pairs: 14 and 44.
Finding the First Digit of the Root:
• Look for a number Y whose square is less than or equal to the first pair (14). The largest such number is 3, because 3² = 9 and 4² = 16 (which is too large).
• Now, divide 14 by 3. The quotient is approximately 3 and the remainder is 14 – 3×3 = 14−9 = 5.
Form the Next Dividend:
• Bring down the next pair (44) next to the remainder 5 to form a new dividend, 544.
Updating the Divisor:
• Add the last digit of the current quotient (3) to the divisor (3), so 3+3 = 6.
• Find the largest digit Z such that when added to 6 as the tens place, and multiplied by itself, it is less than or equal to 544. The correct Z here is 8, because 68×8 = 544.
Perform the Division:
• Divide 544 by 68, resulting in a quotient of 8 and a remainder of 544−68×8 = 544−544 = 0.
Conclusion:
• Since the remainder is 0 and there are no more digits to bring down, the process stops here.
• The quotient, 38, is the square root of 1444.
## 1444 is Perfect Square root or Not
Yes, 1444 is a Perfect Square.
This calculation shows that 1444 is the result of squaring 38, thereby confirming it as a perfect square. This concept is used in various mathematical contexts, particularly in algebra and geometry, where recognizing perfect squares can help simplify computations and solve equations more effectively. Additionally, perfect squares play a role in number theory and are essential for understanding the properties of integers and their relationships.
## Why is understanding squares and square roots important in mathematics?
Understanding squares and square roots is crucial in mathematics for solving equations, understanding geometric concepts like area and volume, and analyzing patterns and relationships in numbers.
## What are the applications of knowing the square and square root of numbers like 1444 in real life?
The Babylonian method, also known as the method of guess and check, is an ancient technique for finding square roots. It involves making an initial guess at the square root, then refining that guess by averaging the guess and the result of dividing the original number by the guess. For 1444, you would start with an initial guess (such as 37 or 39), compute 1444 divided by the guess, and average this quotient with the guess to get a new guess. Repeat until the guesses converge to 38.
Text prompt |
Question
# A motor boat whose speed is 18 Km/h in still water takes 1 hour more to go 24 Km upstream than to return downstream to the same spot. Find the speed of the stream.
Solution
## Given:- Speed of boat $$= 18 \; {km}/{hr}$$Distance $$= 24 \; km$$Let $$x$$ be the speed of stream.Let $${t}_{1}$$ and $${t}_{2}$$ be the time for upstream and downstream.As we know that,$$\text{speed} = \cfrac{\text{distance}}{\text{time}}$$$$\Rightarrow \text{time} = \cfrac{\text{distance}}{\text{speed}}$$For upstream,Speed $$= \left( 18 - x \right) \; {km}/{hr}$$Distance $$= 24 \; km$$Time $$= {t}_{1}$$Therefore,$${t}_{1} = \cfrac{24}{18 - x}$$For downstream,Speed $$= \left( 18 + x \right) \; {km}/{hr}$$Distance $$= 24 \; km$$Time $$= {t}_{2}$$Therefore,$${t}_{2} = \cfrac{24}{18 + x}$$Now according to the question-$${t}_{1} = {t}_{2} + 1$$$$\cfrac{24}{18 - x} = \cfrac{24}{18 + x} + 1$$$$\Rightarrow \cfrac{1}{18 - x} - \cfrac{1}{18 + x} = \cfrac{1}{24}$$$$\Rightarrow \cfrac{\left( 18 + x \right) - \left( 18 - x \right)}{\left( 18 - x \right) \left( 18 + x \right)} = \cfrac{1}{24}$$$$\Rightarrow 48x = \left( 18 - x \right) \left( 18 + x \right)$$$$\Rightarrow 48x = 324 + 18x - 18x - {x}^{2}$$$$\Rightarrow {x}^{2} + 48x - 324 = 0$$$$\Rightarrow {x}^{2} + 54x - 6x - 324 = 0$$$$\Rightarrow x \left( x + 54 \right) - 6 \left( x + 54 \right) = 0$$$$\Rightarrow \left( x + 54 \right) \left( x - 6 \right) = 0$$$$\Rightarrow x = -54 \text{ or } x = 6$$Since speed cannot be negative.$$\Rightarrow x \ne -54$$$$\therefore x = 6$$Thus the speed of stream is $$6 \; {km}/{hr}$$Hence the correct answer is $$6 \; {km}/{hr}$$.Maths
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# Sample Mean
You are about to finish high school, and you have decided it is time for a change of scenery, so you want to go to a university in another city, let's say San Francisco, California. Among your considerations are, how much will I pay for the rent of an apartment, or how much will I spend on public transportation? So, you decide to ask some of your acquaintances who live over there to see how much they spend on average.
This process is called taking a sample mean and in this article you will find the definition, how to calculate a sample mean, standard deviation, variance, the sampling distribution and examples.
## Definition of Sample Means
The mean of a set of numbers is just the average, that is, the sum of all the elements in the set divided by the number of elements in the set.
The sample mean is the average of the values obtained in the sample.
It is easy to see that if two sets are different, they will most likely also have different means.
## Calculation of Sample Means
The sample mean is denoted by $$\overline{x}$$, and is calculated by adding up all the values obtained from the sample and dividing by the total sample size $$n$$. The process is the same as averaging a data set. Therefore, the formula is $\overline{x}=\frac{x_1+\ldots+x_n}{n},$
where $$\overline{x}$$ is the sample mean, $$x_i$$ is each element in the sample and $$n$$ is the sample size.
Let's go back to the San Francisco example. Suppose you asked $$5$$ of your acquaintances how much they spend on public transport per week, and they said $$\20$$, $$\25$$, $$\27$$, $$\43$$, and $$\50$$. So, the sample mean is calculated by:
$\overline{x}=\frac{20+25+27+43+50}{5}=\frac{165}{5}=33.$
Therefore, for this sample, the average amount spent on public transportation in a week is $$33$$.
## Standard Deviation and Variance of the Sample Mean
Since the variance is the square of the standard deviation, to calculate either value, two cases must be considered:
1. You know the population standard deviation.
2. You do not know the population standard deviation.
The following section shows how to calculate this value for each case.
## The Mean and Standard Deviation Formula for Sample Means
The mean of the sample mean, denoted by $$\mu_\overline{x}$$, is given by the population mean, that is if $$\mu$$ is the population mean, $\mu_\overline{x}=\mu.$
To calculate the standard deviation of the sample mean (also called the standard error of the mean (SEM)), denoted by $$\sigma_\overline{x}$$, the two previous cases must be considered. Let's explore them in turn.
### Calculating the Sample Mean Standard Deviation using the Population Standard Deviation
If the sample of size $$n$$ is drawn from a population whose standard deviation $$\sigma$$ is known, then the standard deviation of the sample mean will be given by $\sigma_\overline{x}=\frac{\sigma}{\sqrt{n}}.$
A sample of $$81$$ people was taken from a population with standard deviation $$45$$, what is the standard deviation of the sample mean?
Solution:
Using the formula stated before, the standard deviation of the sample mean is $\sigma_\overline{x}=\frac{45}{\sqrt{81}}=\frac{45}{9}=5.$
Note that to calculate this, you do not need to know anything about the sample besides its size.
### Calculating the Sample Mean Standard Deviation without using the Population Standard Deviation
Sometimes, when you want to estimate the mean of a population, you do not have any information other than just the data from the sample you took. Fortunately, if the sample is large enough (greater than $$30$$), the standard deviation of the sample mean can be approximated using the sample standard deviation. Thus, for a sample of size $$n$$, the standard deviation of the sample mean is $\sigma_\overline{x}\approx\frac{s}{\sqrt{n}},$ where $$s$$ is the sample standard deviation (see the article Standard Deviation for more information) calculated by:
$s=\sqrt{\frac{(x_1-\overline{x})^2+\ldots+(x_n-\overline{x})^2}{n-1}},$
where $$x_i$$ is each element in the sample and $$\overline{x}$$ is the sample mean.
❗❗ The sample standard deviation measures the dispersion of data within the sample, while the sample mean standard deviation measures the dispersion between the means from different samples.
## Sampling Distribution of the Mean
Recall the sampling distribution definition.
The distribution of the sample mean (or sampling distribution of the mean) is the distribution obtained by considering all the means that can be obtained from fixed-size samples in a population.
If $$\overline{x}$$ is the sample mean of a sample of size $$n$$ from a population with mean $$\mu$$ and standard deviation $$\sigma$$. Then, the sampling distribution of $$\overline{x}$$ has mean and standard deviation given by $\mu_\overline{x}=\mu\,\text{ and }\,\sigma_\overline{x}=\frac{\sigma}{\sqrt{n}}.$
Furthermore, if the distribution of the population is normal or the sample size is large enough (according to the Central Limit Theorem, $$n\geq 30$$ is enough), then the sampling distribution of $$\overline{x}$$ is also normal.
When the distribution is normal, you can calculate probabilities using the standard normal distribution table, for this you need to convert the sample mean $$\overline{x}$$ into a $$z$$-score using the following formula
$z=\frac{\overline{x}-\mu_\overline{x}}{\sigma_\overline{x}}=\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}.$
You may be wondering, what happens when the population distribution is not normal and the sample size is small? Unfortunately, for those cases, there is no general way to obtain the shape of the sampling distribution.
Let's see an example of a graph of a sampling distribution of the mean.
Going back to the example of public transportation in San Francisco, let's suppose you had managed to survey thousands of people, grouped the people into groups of size $$10$$, averaged them in each group and obtained the following graph.
Figure 1. Relative frenquency histogram of 360 sample means for the public transport example
This graph approximates the graph of the sampling distribution of the mean. Based on the graph, you can deduce that an average of $$\37$$ is spent on public transportation in San Francisco.
## Examples of Sample Means
Let's see an example of how to calculate probabilities.
It is assumed that the human body temperature distribution has a mean of $$98.6\, °F$$ with a standard deviation of $$2\, °F$$. If a sample of $$49$$ people are taken at random, calculate the following probabilities:
(a) the average temperature of the sample is less than $$98$$, that is, $$P(\overline{x}<98)$$.
(b) the average temperature of the sample is greater than $$99$$, that is, $$P(\overline{x}>99)$$.
(c) the average temperature is between $$98$$ and $$99$$, that is, $$P(98<\overline{x}<99)$$.
Solution:
1. Since the sample size is $$n=49>30$$, you can assume the sampling distribution is normal.
2. Calculating the mean and the standard deviation of the sample mean. Using the formulas stated before, $$\mu_\overline{x}=98.6$$ and the standard deviation $$\sigma_\overline{x}=2/\sqrt{49}=2/7$$.
3. Converting the values into $$z-$$scores and using the standard normal table (see the article Standard Normal Distribution for more information), you'll have for (a):
\begin{align} P(\overline{x}<98) &=P\left(z<\frac{98-98.6}{\frac{2}{7}}\right) \\ &= P(z<-2.1) \\ &=0.0179. \end{align}
For (b) you'll have:
\begin{align} P(\overline{x}>99) &=P\left(z>\frac{99-98.6}{\frac{2}{7}}\right) \\ &= P(z>1.4) \\ &=1-P(z<1.4) \\ &=1-0.9192 \\ &= 0.0808. \end{align}
Finally, for (c):
\begin{align} P(98<\overline{x}<99) &=P(\overline{x}<99)-P(\overline{x}<98) \\ &= P(z<1.4)-P(z<-2.1) \\ &= 0.9192-0.0179 \\ &=0.9013. \end{align}
## Sample Mean - Key takeaways
• The sample mean allows you to estimate the population mean.
• The sample mean $$\overline{x}$$ is calculated as an average, that is, $\overline{x}=\frac{x_1+\ldots+x_n}{n},$ where $$x_i$$ is each element in the sample and $$n$$ is the sample size.
• The sampling distribution of the mean $$\overline{x}$$ has mean and standard deviation given by $\mu_\overline{x}=\mu\,\text{ and }\,\sigma_\overline{x}=\frac{\sigma}{\sqrt{n}}.$
• When the sample size is greater than $$30$$, according to the Central Limit Theorem, the sampling distribution of the mean is similar to a normal distribution.
The sample mean is the average of the values obtained in the sample.
By adding up all the values obtained from a sample and dividing by the number of values in the sample.
The formula for calculating the sample mean is (x1+...+xn)/n, where xi is each element in the sample and n is the sample size.
The most obvious benefit of computing the sample mean is that it provides reliable information that can be applied to the bigger group/population. This is significant since it allows for statistical analysis without the impossibility of polling every person involved.
The main disadvantage is that you cannot find extreme values, either very high or very low, since taking the average of them makes you get a value close to the mean. Another disadvantage is that it is sometimes difficult to select good samples, so there is a possibility of getting biased answers.
## Final Sample Mean Quiz
Question
If you have information about the population, which formula will you use to calculate the standard deviation of a sample mean $$\overline{x}$$?
$$\sigma_\overline{x}=\frac{s}{\sqrt{n}}$$.
Show question
Question
If you don't have information about the population, which formula will you use to calculate the standard deviation of a sample mean $$\overline{x}$$?
$$\sigma_\overline{x}=\frac{s}{\sqrt{n}}$$.
Show question
Question
How do you calculate the sample mean?
$$\overline{x}=\frac{x_1+\ldots+x_n}{n}.$$
Show question
Question
What is the formula for converting a sample mean into a $$z$$-score?
$$z=\frac{\overline{x}-\mu_\overline{x}}{\sigma_\overline{x}}$$.
Show question
Question
The distribution of the sample mean can be normal even if the distribution of the population is not normal.
True.
Show question
Question
Which condition regarding the sample size must be met for the sampling distribution of the mean to be normal?
$$n\geq 30$$.
Show question
Question
Explain the difference between $$\mu$$ and $$\mu_\overline{x}$$.
$$\mu$$ is the population mean, while $$\mu_\overline{x}$$ is the mean of the sampling distribution of $$\overline{x}$$.
Show question
Question
From a population with mean $$\mu=5$$ and standard deviation $$\sigma=2$$, a sample of size $$25$$ is taken. What is the standard deviation of the sample mean?
$$\frac{2}{25}$$.
Show question
Question
From a population with mean $$\mu=10$$ and standard deviation $$\sigma=5$$, a sample of size $$100$$ is taken. What is the standard deviation of the sample mean?
$$\frac{1}{2}$$.
Show question
Question
A population has mean $$\mu$$ and standard deviation $$\sigma$$. For which sample size will the standard deviation of the sample mean be larger?
$$n=10$$.
Show question
Question
A population has mean $$\mu$$ and standard deviation $$\sigma$$. For which sample size will the standard deviation of the sample mean be smaller?
$$n=10$$.
Show question
Question
What does the Central Limit Theorem say about the sampling distribution of the mean?
If $$n\geq 30$$ then, the sampling distribution of the mean is approximately normal.
Show question
Question
If the sampling distribution of the mean $$\overline{x}$$ is approximately normal, has mean $$25$$ and standard deviation $$11$$, how do you calculate the probability $$P(\overline{x}<23)$$?
$$P(z<\frac{2}{11})$$.
Show question
Question
If the sampling distribution of the mean $$\overline{x}$$ is approximately normal, has mean $$58$$ and standard deviation $$3$$. What is the formula for converting a value $$\overline{x}$$ into a $$z$$-score?
$$z=\frac{\overline{x}-58}{3}$$.
Show question
Question
Explain the difference between $$\sigma$$ and $$\sigma_\overline{x}$$.
$$\sigma$$ is the standard deviation of the population, while $$\sigma_\overline{x}$$ is the standard deviation of the sampling distribution of $$\overline{x}$$.
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# Question 50f95
Jun 10, 2015
Unit conversions with scientific notation are all about base 10 exponents.
#### Explanation:
I'll show you how to do a couple of conversions using scientific notation and then redirect you to some videos.
When you're doing unit conversion with scientific notation, you have to keep track of the base 10 exponents. If you're converting between two SI units, the base 10 exponents will be the only ones that change.
Until you get used to the multiplication factors that get you from one unit to another, use a metric conversion chart.
So, in your case, you have $1.78 \cdot {10}^{6} \text{g}$. Let's say that you want to convert this value to kilograms. The chart tells you that you need a multiplication factor of ${10}^{3}$ to get from the unit, in your case grams, to the kilo unit, in your ase kilograms.
This means that you need 1000 g to make a kg, or, in other words, that 1 kg contains ${10}^{3}$ **grams. You would write
1.78 * 10^(6)cancel("g") * "1 kg"/(10^(3)cancel("g")) = (1.78 * 10^cancel(6))/cancel(10^(3)) = 1.78 * 10^(3)"kg"
Notice that you only had to work with the base 10 exponents.
Now let's say that you want to convert this value to micrograms. The multiplication factor is equal to ${10}^{- 6}$, which means that you need ${10}^{6}$ micrograms to get 1 gram.
You would write
1.78 * 10^(6)cancel("g") * (10^(6)mu"g")/(1cancel("g")) = 1.78 * 10^(6 + 6) = 1.78 * 10^(12)mu"g"
Now let's say that you want to convert to gigagrams. The multiplication factor is equal to ${10}^{9}$, so you need ${10}^{9}$ grams to get to 1 gigagram. You would write
1.78*10^6cancel("g") * "1 Gg"/(10^(9)cancel("g")) = (1.78 * cancel(10^(6)))/10^(cancel(9)) = 1.78/10^3 = 1.78 * 10^(-3)"Gg"
Now, let's say that you want to convert to ounces. The conversion factor between these two units has 1 ounce equal to 28.3495231 grams. This time, the base 10 exponent is not the only one that changes. You would write
1.78*color(blue)(10^(6))cancel("g") * "1 ounce"/(28.3495231cancel("g")) = 0.0628 * color(blue)(10^(6)) = 6.28 * 10^(4)"ounces"#
Two things to notice here
• You divide the regular numbers first, and leave the base 10 number unchanged;
• Then you use the base 10 number to express the result in scientific notation.
You can check out the videos available on Socratic:
http://socratic.org/chemistry/measurement-in-chemistry/unit-conversions/videos
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# Row Matrix
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Last updated date: 11th Aug 2024
Total views: 201.9k
Views today: 3.01k
## An Introduction to the Concept of Row Matrix
A row Matrix is just a mathematical way of representing a list of numbers in horizontal form. We may encounter different sets of data which can be represented in a list or an array form, such as the price of different stationary items or marks got by a student in different subjects and many more. Writing this data in the form of a Row Matrix makes it simple to read and understand the data.
## What is Row Matrix?
A Row Matrix is a matrix having a single row. Recall that a Matrix is a rectangular array of numbers or symbols which is used to represent some physical object, having numbers arranged in rows and columns. In mathematical form, A Matrix having m rows and n columns is represented by
$A = {\left[ {{a_{ij}}} \right]_{m \times n}}$
So, a Row Matrix will have only one row and n columns. Hence, a Row Matrix is represented by ${[{a_{1j}}]_{1 \times n}}$where ${a_{1j}}$is the (1,j)th entry of the matrix.
## Representation and Size
A Row Matrix is written in the form:
${[{a_{1j}}]_{1 \times n}}$where ${a_{1j}}$is the (1,j)th entry of the matrix. The number of entries of the matrix $\left[a_{1j}\right]_{1 \times n}$ is n. The number of entities in a row matrix depends on the number of columns. In other words, the number of entities of a row matrix is equal to the number of columns. A Row Matrix always has only one row. It can have any number of columns. We say that a row Matrix has n columns. So, a Row Matrix is n-dimensional.
## Examples of Row Matrix
• A Row Matrix of order 1×2 is $\left[ {\begin{array}{*{20}{c}}2&5\end{array}} \right]$.
• A Row Matrix of size 5 is $\left[ {\begin{array}{*{20}{c}}{2.5}&{3.6}&{5.1}&{ - 2.0}&{6.3}\end{array}} \right]$.
• Row Matrix containing 3 elements is $\left[ {\begin{array}{*{20}{c}}{60}&{40}&{50}\end{array}} \right]$.
• Row Matrix of order 1×4 with alternative entries 1 and 0 is $\left[ {\begin{array}{*{20}{c}}1&0&1&0\end{array}} \right]$.
## Practical Uses of Row Matrix
• While sending a message over a network, the sender device encodes it into a binary format using an array of 0's and 1's in Row Matrix Form. This is known as Cryptography.
• We can store data such as the Family's total monthly cost using a one-dimensional horizontal matrix, i.e., a Row Matrix.
## Properties of Row Matrix
• Commutativity
• $A + B = B + A$
• Associativity
• $A + \left( {B + C} \right) = \left( {A + B} \right) + C$
## Operations on Row Matrix
### 1. Addition of Row Matrices
We can add two row matrices by simply adding their corresponding entries.
This is depicted as:
Given two matrices $A = \left[ {\begin{array}{*{20}{c}}1&2&3\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}2&3&4\end{array}} \right]$, We need to find $C = A + B$. Implies
$\begin{array}{l}C = \left[ {\begin{array}{*{20}{c}}1&2&3\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}2&3&4\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1 + 2}&{2 + 3}&{3 + 4}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}3&5&7\end{array}} \right]\end{array}$
So, the addition of two matrices is $\left[ {\begin{array}{*{20}{c}}3&5&7\end{array}} \right]$.
Similarly, we can add three matrices together by adding their corresponding entries as:
Given $P = \left[ {\begin{array}{*{20}{c}}2&{2.5}&3&{1.5}\end{array}} \right]$, $Q = \left[ {\begin{array}{*{20}{c}}{1.5}&{ - 1}&{ - 4.25}&5\end{array}} \right]$, $R = \left[ {\begin{array}{*{20}{c}}0&{ - 1.5}&1&{ - 2.5}\end{array}} \right]$. We need to find $S = P + Q + R$. Putting values of P, Q, and R, we get
$S = \left[ {\begin{array}{*{20}{c}}2&{2.5}&3&{1.5}\end{array}\left] + \right[\begin{array}{*{20}{c}}{1.5\;\,}&{ - 1}&{ - 4.25}&5\end{array}\left] + \right[\begin{array}{*{20}{c}}{0\;}&{ - 1.5}&1&{ - 2.5}\end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}}{2 + 1.5 + 0\;}&{2.5 + \left( { - 1} \right) + \left( { - 1.5} \right)}&{3 + \left( { - 4.25} \right) + 1\;}&{1.5 + 5 + \left( { - 2.5} \right)}\end{array}\;} \right]$
$= {\rm{ }}\left[ {\begin{array}{*{20}{c}}{3.5}&0&{ - 0.25}&{4.0}\end{array}} \right]$
### 2. Subtraction of Row Matrices
We can subtract one matrix from another by subtracting their corresponding ith entries. This is depicted as:
Given two matrices $A = \left[ {\begin{array}{*{20}{c}}1&2&3\end{array}} \right]$ and$B = \left[ {\begin{array}{*{20}{c}}2&3&4\end{array}} \right]$. We need to find $D = A - B$. Implies
$\begin{array}{l}D = \left[ {\begin{array}{*{20}{c}}1&2&3\end{array}} \right]{\rm{ }} - \left[ {\begin{array}{*{20}{c}}2&3&4\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1 - 2}&{2 - 3}&{3 - 4}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 1}&{ - 1}\end{array}} \right]\end{array}$
So, the subtraction of second matrix from first is $\left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 1}&{ - 1}\end{array}} \right]$.
Similarly, we can subtract two matrices from a given matrix together as:
Given $P = \left[ {\begin{array}{*{20}{c}}2&{2.5}&3&{1.5}\end{array}} \right]$, $Q = \left[ {\begin{array}{*{20}{c}}{1.5}&{ - 1}&{ - 4.25}&5\end{array}} \right]$, $R = \left[ {\begin{array}{*{20}{c}}0&{ - 1.5}&1&{ - 2.5}\end{array}} \right]$. We need $T = P - Q - R$. Putting values of P, Q, and R, we get
$\begin{array}{l}S = \left[ {\begin{array}{*{20}{c}}2&{2.5}&3&{1.5}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{1.5}&{ - 1}&{ - 4.25}&5\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}0&{ - 1.5}&1&{ - 2.5}\end{array}} \right]{\rm{ }}\\ = \left[ {\begin{array}{*{20}{c}}{2 - 1.5 - 0\;}&{2.5 - \left( { - 1} \right) - \left( { - 1.5} \right)}&{3 - \left( { - 4.25} \right) - 1\;}&{1.5 - 5 - \left( { - 2.5} \right)}\end{array}\;} \right]\\ = \left[ {\begin{array}{*{20}{c}}{0.5}&{5.0}&{6.25}&{ - 1}\end{array}{\rm{.0}}} \right]\end{array}$.
This can also be done as $T = P - \left( {Q + R} \right)$.
Put $Q + R = M$, we get
$\begin{array}{*{20}{l}}{M = \left[ {\begin{array}{*{20}{c}}{1.5 + 0}&{ - 1 + \left( { - 1.5} \right)}&{\left( { - 4.25} \right) + 1}&{{\rm{ }}5 + \left( { - 2.5} \right)}\end{array}} \right]}\\{ = \left[ {\begin{array}{*{20}{c}}{1.5}&{ - 2.5}&{ - 3.25}&{2.5}\end{array}} \right]}\end{array}$
Now,
$\begin{array}{*{20}{l}}{S = P - M}\\\begin{array}{l} = \left[ {\begin{array}{*{20}{c}}2&{2.5}&3&{1.5}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{1.5}&{ - 2.5}&{ - 3.25}&{2.5}\end{array}} \right]{\rm{ }}\\ = \left[ {\begin{array}{*{20}{c}}{0.5}&{5.0}&{6.25}&{ - 1.0}\end{array}} \right]\end{array}\end{array}$
### 3. Multiplication of Row Matrix by a Scalar
We can multiply a matrix by a scalar as: Given $B = [\begin{array}{*{20}{c}}{{b_{11}}}&{{b_{12}}}&{{b_{13}}}& \ldots &{{b_{1n}}}\end{array}]$. Multiplying B by a scalar k, we get $kB = k[\begin{array}{*{20}{c}}{{b_{11}}}&{{b_{12}}}&{{b_{13}}}& \ldots &{{b_{1n}}}\end{array}]$
$kB = [\begin{array}{*{20}{c}}{k{b_{11}}}&{k{b_{12}}}&{k{b_{13}}}& \ldots &{k{b_{1n}}}\end{array}]$.
For example:
$\begin{array}{*{20}{l}}\begin{array}{l}A = {\left[ {\begin{array}{*{20}{c}}2&{ - 1}&3\end{array}} \right]_{1 \times 3}},k = 2,{\rm{ }}\\kA = 2A = 2\left[ {\begin{array}{*{20}{c}}2&{ - 1}&3\end{array}} \right]\end{array}\\\begin{array}{l} = \left[ {\begin{array}{*{20}{c}}{2(2)}&{2( - 1)}&{2(3)}\end{array}} \right]{\rm{ }}\\ = \left[ {\begin{array}{*{20}{c}}4&{ - 2}&6\end{array}} \right]\end{array}\end{array}$
## Solved Questions
1. Find the value of${\bf{A}} - {\bf{B}} + {\bf{2C}}$, where ${\bf{A}} = \left[ {\begin{array}{*{20}{c}}1&2\end{array}} \right],{\rm{ }}{\bf{B}} = \left[ {\begin{array}{*{20}{c}}3&4\end{array}} \right],{\rm{ }}{\bf{C}} = \left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 2}\end{array}} \right]$.
$\begin{array}{*{20}{l}}{Let {\rm{ }}S = A - B + 2C}\\{ = \left[ {\begin{array}{*{20}{c}}1&2\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}3&4\end{array}} \right] + 2\left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 2}\end{array}} \right]}\\\begin{array}{l} = \left[ {\begin{array}{*{20}{c}}1&2\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}3&4\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 2}&{ - 4}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1 - 3 - 2}&{2 - 4 - 4}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 4}&{ - 6}\end{array}} \right]\end{array}\end{array}$
2. Given${\bf{A}} = \left[ {\begin{array}{*{20}{c}}2&3&5\end{array}} \right],{\rm{ }}{\bf{B}} = \left[ {\begin{array}{*{20}{c}}{ - 1}&5&6\end{array}} \right],{\rm{ }}{\bf{M}} = \left[ {\begin{array}{*{20}{c}}{ - 4}&2&{ - 5}\end{array}} \right]$. Find Matrix C for${\bf{M}} = {\bf{C}}+{\bf{3A}} + {\bf{2B}}$.
Given ${\bf{M}} = {\bf{3A}} + {\bf{2B}} +{\bf{C}}$. Solving for C, we get $C = M - 3A - 2B$. Putting values of A, B, and M, we get
$\begin{array}{*{20}{l}}{C = \left[ {\begin{array}{*{20}{c}}{ - 4}&2&{ - 5}\end{array}} \right] - 3\left[ {\begin{array}{*{20}{c}}2&3&5\end{array}} \right] - 2\left[ {\begin{array}{*{20}{c}}{ - 1}&5&6\end{array}} \right]}\\\begin{array}{l} = \left[ {\begin{array}{*{20}{c}}{ - 4}&2&{ - 5}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}6&9&{15}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{ - 2}&{10}&{12}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 4 - 6 + 2}&{2 - 9 - 10}&{ - 5 - 15 - 12}\end{array}} \right]{\rm{ }}\\ = \left[ {\begin{array}{*{20}{c}}{ - 8}&{ - 17}&{ - 32}\end{array}} \right]\end{array}\end{array}$
## Practice Questions
1. If $A = \left[ {\begin{array}{*{20}{c}}2&3&x\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{c}}y&3&5\end{array}} \right]$ and A = B, then find the value of x and y.
Answer: x = 5, y = 2
2. If A and B two row matrices and AB exist, then find the number of columns of A.
Answer: The number of columns of A is 1.
3. Why does your Word document need to be formatted?
## Conclusion
A Row Matrix is a horizontal matrix. It can also be called an array. Two or more Row Matrices can be added or subtracted if the order of both matrices is the same. We can multiply a scalar with a row matrix. Row matrix follows associativity property.
Competitive Exams after 12th Science
## FAQs on Row Matrix
1. What is the condition to multiply two matrices?
The condition is the number of columns of the first matrix is equal to the number of rows of the second matrix.
2. When is the matrix multiplication commutative?
The matrix multiplication does not follow the commutative property.
But $I \cdot A =A \cdot I$, where I is an identity matrix and A is a square matrix. Matrix multiplication is commutative if a square matrix multiplied with an identity matrix.
3. What is the order of the matrix that is produced by multiplying two matrices with order $m\times n$ and $n\times p$?
The number of rows of the product matrix is the number of rows of the first matrix. The number of columns of the product matrix is the number of columns of the second matrix. The order of the product matrix is $m \times p$. |
# Area of a Kite Calculator
When you think about a kite in mathematics, it is not the same kite you regularly see in your life. In our daily lives, we see multiple kites having different shapes and dimensions. In Mathematics, the kite refers to a 2D figure having a specific figure similar to a cone from the bottom. Finding its area can be a complex process because of its complex shape.
If you don’t want to get involved manually, you should use this area of a kite calculator. With the help of this maths calculator, you can easily calculate the area of a kite. It is accessible to every mathematics student because of its simple interface and efficient work. You will not find any mistake in this tool’s solution that makes it suitable for solving all types of assignments.
## Area of a Kite Definition
To understand what is the area of a kite, you should first learn a little about this figure. A kite is a 2D plane figure which has four total sides. It can be divided into two main sections that are upper section and the lower section. The two opposite sides in the upper part are the same and the two opposite sides in the lower portion are the same in length.
It looks like a cone from the bottom part while the triangular cap from the upper side. In Mathematics, the area of a kite is the region in space that it covers. When a kite is placed in space or on a solid surface, the region covered by this figure is called its area. Calculating this measurement can either be hectic or simple just according to the understanding you have with this figure.
### Area of a Kite Formula
As mentioned earlier, the opposite sides of a kite are the same in length. So, you might be thinking that calculating area might be similar to that of a rectangle because of this common property. But it is not right as it has a different shape.
Therefore, you need to use another formula for the area calculation of a kite. It involves the distance between the upper and lower corner of the kite as well as the distance between the left and right corners of the kite. Here is the formula you need to use for calculating the area of a kite.
Area of a kite = (e x f) /2 square units
Here:
• “e” shows the distance between the left and right corners (horizontal diagonal) of the kite available in the upper part.
• “f” represents the distance between the upper and lower corners (vertical diagonal) of the kite and it is calculated vertically.
### How to calculate the area of a kite?
With the above formula, you must have got an idea of how to calculate the area of a kite. To let you understand properly, we have enlisted some points here and then solved an example too. Let’s have a look at both of these for a better understanding.
• Measure the distance between the left and right corners of the kite from the top portion.
• Measure the distance between the upper and lower corners of the kite.
• Multiply these measures.
• Divide the final answer by “2” to get the final answer.
Keep in mind that the units of the area of a kite must be square of the given lengths units. For example, if you have measurements in meters, the units for the area of a kite will be “meter2 or m2”.
Here is a solved example for your better understanding.
Example 1:
Find the area of a kite if its horizontal diagonal is 3 meters long and its vertical diagonal is 8 meters long.
Solution:
As we know,
Area of a kite = (e x f) /2 square units
So,
= (3 x 8) /2 m2
= 24/2 m2
Area = 12 m2
### How to use the area of a kite?
If you have large dimensions for a kite or are unable to find the area of a kite manually, you should use this online tool by Calculator’s Bag. You can use it simply just by giving input values you have and get the final answer.
• Insert the measure for the horizontal diagonal
• Put the measure for the vertical diagonal
• This calculator will automatically show the answer for the area of a kite.
### FAQ | Area of a kite
How do you calculate the area of a kite?
To calculate the area of a kite, you need to multiply the horizontal and vertical diagonals. Then, divide the final answer by “2” to get the area of that specific kite.
How to find the area of the kite?
To find the area of the kite, you need to use the following formula:
Area of a kite = (e x f) /2 square units
What is the area and perimeter of a kite?
The area of a kite is calculated by multiplying the diagonals and then dividing the answer by “2”. On the other side, you have to add all four sides to find the perimeter of the kite.
Is the area of a rhombus the same as a kite?
Yes, a rhombus is much similar in shape to a kite. That’s why, the area of a rhombus is the same as a kite.
Do angles in a kite add up to 360?
Yes, the interior angles of a kite will make 360 degrees when summed up.
Is every kite a square?
No, every kite is not a square.
What are the angle rules for a kite?
• The sum of all angles will be 360 degrees.
• The angles, where two different sides meet, will be equal to each other. |
Question 1: Compute:
i) $\frac{30!}{28!}$ ii) $\frac{11!-10!}{9!}$ iii) LCM $( 6!, 7!, 8!)$
i) $\frac{30!}{28!}$ $=$ $\frac{30 \times 29 \times 28!}{28!}$ $= 30 \times 29 = 870$
ii) $\frac{11!-10!}{9!}$ $=$ $\frac{11 \times 10 \times 9! - 10 \times 9!}{9!}$ $= 11 \times 10 - 10 = 100$
iii) LCM $( 6!, 7!, 8!)$
$8! = 8 \times 7 \times 6!$
$7! = 7 \times 6!$
$6! = 6!$
Therefore LCM $( 6!, 7!, 8!)$ is $8 \times 7 \times 6! = 8!$
$\\$
Question 2: Prove that $\frac{1}{9!}$ $+$ $\frac{1}{10!}$ $+$ $\frac{1}{11!}$ $=$ $\frac{122}{11!}$
LHS $=$ $\frac{1}{9!}$ $+$ $\frac{1}{10!}$ $+$ $\frac{1}{11!}$
$=$ $\frac{1}{9!}$ $+$ $\frac{1}{10 \times 9!}$ $+$ $\frac{1}{11 \times 10 \times 9!}$
$=$ $\frac{1}{9!}$ $\Big( 1 +$ $\frac{1}{10}$ $+$ $\frac{1}{110}$ $\Big)$
$=$ $\frac{1}{9!}$ $\Big($ $\frac{110+11+1}{110}$ $\Big)$
$=$ $\frac{122}{11!}$ $=$ RHS. Hence proved.
$\\$
Question 3: Find x in each of the following:
i) $\frac{1}{4!}$ $+$ $\frac{1}{5!}$ $=$ $\frac{x}{6!}$ ii) $\frac{x}{10!}$ $=$ $\frac{1}{8!}$ $+$ $\frac{1}{9!}$ iii) $\frac{1}{6!}$ $+$ $\frac{1}{7!}$ $=$ $\frac{x}{8!}$
i) $\frac{1}{4!}$ $+$ $\frac{1}{5!}$ $=$ $\frac{x}{6!}$
$\Rightarrow$ $\frac{1}{4!}$ $+$ $\frac{1}{5 \times 4!}$ $=$ $\frac{x}{6 \times 5 \times 4!}$
$\Rightarrow 1 +$ $\frac{1}{5 }$ $=$ $\frac{x}{6 \times 5 }$
$\Rightarrow$ $\frac{6}{5}$ $=$ $\frac{x}{30}$
$\Rightarrow x= 36$
ii) $\frac{x}{10!}$ $=$ $\frac{1}{8!}$ $+$ $\frac{1}{9!}$
$\Rightarrow$ $\frac{x}{10 \times 9 \times 8!}$ $=$ $\frac{1}{8!}$ $+$ $\frac{1}{9 \times 8!}$
$\Rightarrow$ $\frac{x}{90}$ $= 1+$ $\frac{1}{9}$
$\Rightarrow$ $\frac{x}{90}$ $=$ $\frac{10}{9}$
$\Rightarrow x = 100$
iii) $\frac{1}{6!}$ $+$ $\frac{1}{7!}$ $=$ $\frac{x}{8!}$
$\Rightarrow$ $\frac{1}{6!}$ $+$ $\frac{1}{7 \times 6!}$ $=$ $\frac{x}{8\times 7 \times 6!}$
$\Rightarrow 1 +$ $\frac{1}{7}$ $=$ $\frac{x}{56}$
$\Rightarrow$ $\frac{8}{7}$ $=$ $\frac{x}{56}$
$\Rightarrow x = 64$
$\\$
Question 4: Convert the following products into factorials:
i) $5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10$ ii) $3 \cdot 6 \cdot 9 \cdot 12 \cdot 15 \cdot 18$
iii) $(n+1)(n+2)(n+3) \ldots (2n)$ iv) $1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \ldots (2n-1)$
i) $5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10 =$ $\frac{(1 \times 2 \times 3 \times 4 )( 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10)}{1 \times 2 \times 3 \times 4 }$ $=$ $\frac{10!}{4!}$
ii) $3 \cdot 6 \cdot 9 \cdot 12 \cdot 15 \cdot 18 = 3^6 ( 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6) = 3^6 \times 6!$
iii) $(n+1)(n+2)(n+3) \ldots (2n) =$ $\frac{(1 \times 2 \times 3 \ldots \times n)[ (n+1)(n+2)(n+3) \ldots (2n) ] }{(1 \times 2 \times 3 \ldots n)}$ $=$ $\frac{(2n)!}{n!}$
iv) $1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \ldots (2n-1)$
$=$ $\frac{[ 1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \ldots (2n-1) ] [2 \cdot 4 \cdot 6 \cdot 8 \ldots 2n ] }{ [ 2 \cdot 4 \cdot 6 \cdot 8 \ldots 2n ] }$ $=$ $\frac{(2n)!}{2^n ( 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \ldots n)}$ $=$ $\frac{(2n)!}{2^n n!}$
$\\$
Question 5: Which of the following are true:
i) $(2+3)! = 2! + 3!$ ii) $(2 \times 3)! = 2! \times 3!$
i) $(2+3)! = 2! + 3!$
LHS $= ( 2 + 3)! = 5! = 120$
RHS $= 2! + 3! = 2 + 6 = 8$
Hence LHS $\neq$ RHS. Hence the statement is false.
ii) $(2 \times 3)! = 2! \times 3!$
LHS $= (2 \times 3)! = 6! = 720$
RHS $= 2! \times 3! = 2 \times 6 = 12$
Hence LHS $\neq$ RHS. Hence the statement is false.
$\\$
Question 6: Prove that: $n! ( n+2) = n! + ( n+1)!$
RHS $= n! + ( n+1)! = n! + (n+1) n! = n! ( 1+ n+1) = n! ( n+2) =$ LHS.
Hence proved.
$\\$
Question 7: If $( n+2)! = 60[(n-1)!]$, find $n$
Given $( n+2)! = 60[(n-1)!]$
$\Rightarrow (n+2)(n+1)n(n-1)! = 60(n-1)!$
$\Rightarrow (n+2)(n+1)n = 60$
$\Rightarrow (n+2)(n+1)n = 5\times 4 \times 3$
$\Rightarrow n = 3$
$\\$
Question 8: If $( n+1)! = 90 [( n-1)!]$, find $n$
Given $( n+1)! = 90 [( n-1)!]$
$\Rightarrow (n+1) \times n \times ( n-1)! = 90 ( n-1)!$
$\Rightarrow n ( n+1) = 90$
$\Rightarrow n ( n+1) = 9 \times 10$
$\Rightarrow n = 9$
$\\$
Question 9: If $( n+3) ! = 56 [(n+1)!]$, find $n$
Given $( n+3) ! = 56 [(n+1)!]$
$\Rightarrow (n+3)(n+2)(n+1)! = 56 (n+1)!$
$\Rightarrow (n+3)(n+2) = 56$
$\Rightarrow (n+3)(n+2) = 8\times 7$
$\Rightarrow n = 5$
$\\$
Question 10: If $\frac{(2n)!}{3! ( 2n-3)!}$ and $\frac{n!}{2! ( n-2)!}$ are in the ratio $44:3$, find $n$
$\frac{(2n)!}{3! ( 2n-3)!}$ $:$ $\frac{n!}{2! ( n-2)!}$ $= 44:3$
$\Rightarrow \frac{(2n)(2n-1)(2n-2)(2n-3)!}{3! ( 2n-3)!}$ $:$ $\frac{n(n-1)(n-2)!}{2! ( n-2)!}$ $= 44:3$
$\Rightarrow \frac{(2n)(2n-1)(2n-2)}{3! }$ $:$ $\frac{n(n-1)}{2!}$ $= 44:3$
$\Rightarrow \frac{(2n)(2n-1)(2n-2)}{3! }$ $\times$ $\frac{2!}{n(n-1)}$ $= \frac{44}{3}$
$\Rightarrow \frac{4(2n-1)}{3}$ $\times$ $\frac{2!}{n(n-1)}$ $=$ $\frac{44}{3}$
$\Rightarrow 2n-1 = 11$
$\Rightarrow 2n = 12$
$\Rightarrow n = 6$
$\\$
Question 11: Prove that:
i) $\frac{n!}{(n-r)!}$ $= n(n-1)(n-2) \ldots (n-(n-r))$
ii) $\frac{n!}{(n-r)!r!}$ $+$ $\frac{n!}{(n-r+1)!(r-1)!}$ $=$ $\frac{(n+1)!}{r!}$
i) $\frac{n!}{(n-r)!}$ $= n(n-1)(n-2) \ldots (n-(n-r))$
LHS $=$ $\frac{n!}{(n-r)!}$ $=$ $\frac{n(n-1)(n-2) \ldots [(n - ( r-1)](n-r)!}{(n-r)!}$
$= n(n-1)(n-2) \ldots [(n - ( r-1)] =$ RHS. Hence proved.
ii) $\frac{n!}{(n-r)!r!}$ $+$ $\frac{n!}{(n-r+1)!(r-1)!}$ $=$ $\frac{(n+1)!}{r!}$
LHS $=$ $\frac{n!}{(n-r)!r!}$ $+$ $\frac{n!}{(n-r+1)!(r-1)!}$
$=$ $\frac{n!}{(n-r)!r(r-1)!}$ $+$ $\frac{n!}{(n-r+1)( n-r)!(r-1)!}$
$=$ $\frac{n!}{(n-r)!(r-1)!}$ $\Big[$ $\frac{1}{r}$ $+$ $\frac{1}{n-r+1}$ $\Big]$
$=$ $\frac{n!}{(n-r)!(r-1)!}$ $\Big[$ $\frac{n - r + 1 + r}{r ( n - r +1)}$ $\Big]$
$=$ $\frac{n!}{(n-r)!(r-1)!}$ $\Big[$ $\frac{n + 1}{r ( n - r +1)}$ $\Big]$
$=$ $\frac{(n+1)!}{r!(n-r)! (n-r+1)!}$
$=$ $\frac{(n+1)!}{(n - r + 1)! r!}$ $=$ RHS.
$\\$
Question 12: Prove that: $\frac{(2n+1)!}{n!}$ $= 2^n \Big\{ 1\cdot 3 \cdot 5 \ldots (2n-1)(2n+1) \Big\}$
LHS $=$ $\frac{(2n+1)!}{n!}$
$=$ $\frac{(2n+1) (2n)!}{n!}$
$=$ $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \ldots n(n+1) \ldots (2n-2) \cdot (2n-1) \cdot (2n) \cdot (2n+1)]}{n!}$
$=$ $\frac{[1 \cdot 3 \cdot 5 \cdot 7 \ldots (2n-1)(2n+1)][2 \cdot 4 \cdot 6 \cdot 8 \ldots (2n-2)(2n)]}{n!}$
$=$ $\frac{[1 \cdot 3 \cdot 5 \cdot 7 \ldots (2n-1)(2n+1)][2^n 1 \cdot 2 \cdot 3 \cdot 4 \ldots (n-1)(n)]}{n!}$
$= 2^n [ 1 \cdot 3 \cdot 5 \cdot 7 \ldots (2n-1)(2n+1)] =$ RHS. Hence proved.
$\\$ |
Ex.13.1 Q2 Surface Areas and Volumes - NCERT Maths Class 9
Go back to 'Ex.13.1'
Question
The length, breadth and height of a room are $$5 \,\rm{m}, 4 \,\rm{m},$$ and $$3\,\rm{ m}$$ respectively. Find the cost of white washing the walls of the room and ceiling at the rate of \begin{align} \text{Rs 7.50 per }\rm{m^2}? \end{align}
Text Solution
What is known:
Medium
What is unknown:
The lenght, breadth and height of a room are $$5m$$, $$4m$$, and $$3m$$ respectively.
Reasoning:
Since the four walls and ceiling are to be whitewashed. So, it has $$5$$ faces only, excluding the base.
Hence, area of the room to be whitewashed can be obtained by adding area of the ceiling to the lateral surface area of the cuboidal room.
Lateral surface area of cuboid $$=2(l+b)h$$
The cost of white washing the walls of the room and ceiling will be equal to area of the room to be whitewashed multiplied by rate of the whitewashing.
Steps:
\begin{align} \\\text{length} &= 5\,\rm{m} \\\text{breadth} &= \rm{4\,m} \\\text{height} &= \rm{3 \,\,m} \end{align}
Surface are of $$5$$ faces $$=$$ Area of the $$4$$ walls and ceiling $$=$$ $$lb+2(l+b)h$$
\begin{align}&=(5m\!\times\!4m)\!+\!2\!\times\!(5m\!+\!4m)\!\times\!3m\\ &= 20m^2+2\times9m\times3m \\&=20m^2+54m^2 \\&=74m^2\end{align}
The cost of white washing the walls of the room and ceiling = $$Rate\times area$$
\begin{align}&=₹ 7.50/m^2\times 74m^2\\ &= ₹\,555\end{align}
Cost of white washing the walls of the room and the ceiling $$= Rs. 555$$ |
Question Video: Finding the Measure of an Angle given Its Supplementary Angle’s Measure | Nagwa Question Video: Finding the Measure of an Angle given Its Supplementary Angle’s Measure | Nagwa
# Question Video: Finding the Measure of an Angle given Its Supplementary Angleβs Measure Mathematics • First Year of Preparatory School
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In the figure, the ray π΅π· bisects β π΄π΅πΈ. What is the πβ π΄π΅π·?
02:15
### Video Transcript
In the figure, the ray from π΅ through π· bisects the angle π΄π΅πΈ. What is the measure of angle π΄π΅π·?
In this question we are given a figure and asked to determine the measure of a given angle using a bisector and a given angle measure.
To answer this question, we can begin by recalling that an angle bisector bisects an angle into two angles with equal measure. In this case, we are told that the ray from π΅ through π· bisects angle π΄π΅πΈ. So we must have that the measure of angle π·π΅πΈ is equal to the measure of angle π΄π΅π·. We can add this information onto the figure.
We can then recall that the measure of a straight angle is 180 degrees. And we can see in the figure that angle π΄π΅πΆ is a straight angle. Hence, the sum of the angle measures that make the straight angle is 180 degrees. We have that 180 degrees is equal to 54 degrees plus the measure of angle π·π΅πΈ plus the measure of angle π΄π΅π·.
Since the measure of angle π·π΅πΈ is equal to the measure of angle π΄π΅π·, we can replace its measure with that of the measure of angle π΄π΅π· to obtain two times the measure of angle π΄π΅π·. We can then subtract 54 degrees from both sides of the equation and evaluate to get that 126 degrees is equal to the two times the measure of angle π΄π΅π·. Finally, we can divide both sides of the equation by two to find that the measure of angle π΄π΅π· is 63 degrees.
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Question Video: Determining the Probability of Intersection of Two Events | Nagwa Question Video: Determining the Probability of Intersection of Two Events | Nagwa
# Question Video: Determining the Probability of Intersection of Two Events Mathematics • Third Year of Preparatory School
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A survey asked 49 people if they had visited any clubs recently. 28 had attended Club A, 38 had attended Club B, and 8 had not been to either club. What is the probability that a random person from the sample attended both clubs?
03:29
### Video Transcript
A survey asked 49 people if they had visited any clubs recently. 28 had attended Club A, 38 had attended Club B, and eight had not been to either club. What is the probability that a random person from the sample attended both clubs?
There are lots of ways to approach this problem, but we are going to look at it through a Venn diagram. The left-hand circle will include all the people that attended Club A. The right-hand circle will include all the people that attended Club B. And the intersection of the two circles will be the people that attended both.
As the Venn diagram will contain all 49 people, outside of the circles will be the eight people that haven’t been to either club. There were 28 people that attended Club A. So we can write the number 28 inside the left-hand circle. 38 people had attended Club B. This means we can write 38 inside the right-hand circle. As eight people had not attended either club, we can write the eight outside of the circles.
You’ll notice at present that there is no number inside the intersection of circle A and circle B. Also when we had the three numbers 28, 38, and eight, we get an answer of 74. However, there were only 49 people in the survey. This means that we need to subtract 49 from 74 as some of the people have been counted twice — the people that attended Club A and Club B. 74 minus 49 is equal to 25. This means that there were 25 people that attended both Club A and Club B.
Whilst we could answer the question “what is the probability that a random person attended both clubs?” from the diagram, it is important that we complete the Venn diagram accurately. If we consider circle A, we know that 25 people attended Club A and Club B. There were 28 people in total that attended Club A. Therefore, the number of people that attended only Club A was three as three plus 25 is equal to 28. We can do the same thing for the 38 people that attended Club B: 38 minus 25 is equal to 13. Therefore, 13 people attended Club B only.
We can check that our diagram is correct by adding the four numbers: 25 plus three plus 13 plus eight. As this is equal to 49, we know that our Venn diagram is correct.
The probability of an event occurring is the number of successful outcomes divided by the number of possible outcomes. In this case, the probability that a person attended both clubs is 25 out of 49 or twenty-five forty-ninths as they were 25 people that attended both clubs and 49 people who took part in the survey in total.
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Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!
• Interactive Sessions
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A cube is a special type of a rectangular solid in which the length, width and height are exactly same. In geometry, a cube is a 3-dimensional solid, bounded by 8 vertices, 6 square faces, 12 sides, among which three meets at each vertex.
## Cube Definition
The definition of a cube states that it is a 3D, six sided solid figure with the same length, breadth, and height. The image given below explains what is a cube.
## Properties of a Cube
A cube is a 3 dimensional object that can be represented on a 2-dimensional surface by using the properties of a cube.
Listed below are some of the properties of a cube:
• All the faces of a cube are squares.
• All the faces meet four other faces.
• All the plane angles are right angles.
• All the vertices meet three faces.
• Opposite edges on faces of a cube are parallel to each other.
• Length of Face diagonal = $\sqrt{2}$ a units, where 'a' is the side of a cube.
• Length of Body diagonal or diagonal of cube = $\sqrt{3}$ a units, where 'a' is the side of a cube.
### Relation between Surface Area and Volume of a Cube:
Volume of a cube = a3 cubic units
Total surface area of a cube = 6a2 square units
Volume = ($\sqrt{\frac{Surface \ Area}{6}})^3$.
## Volume of a Cube
The amount of space occupied by the object is called as cube volume. The unit of measurement for volume of cube is m3, cm3, inches3 etc. The first step to solving problems related to cube is to read and understand what information the question holds. The volume of a cube formula is given as follows:
Volume of a Cube = Side * Side * Side = (Side)3
Given below are some of the examples that explains how to find the volume of a cube with the help of the formula for volume of a cube.
### Solved Example
Question: Find the volume of a cube with side 5 inches.
Solution:
Side of a cube = 5
Volume of a Cube = (Side)3
Volume of a Cube = (5)3
= 5 * 5 * 5
= 125
Therefore, the volume of a cube is 125 cubic inches.
## Surface Area of a Cube
The area of the surfaces of an object is called the surface area of the object. It's unit of measurement is m2, cm2, inches2 etc. The surface area of cube is the area on the outer side of a three-dimensional shape. Surface area formula for a cube helps us to find the length of the outer part of the solid. The formula for the surface area of a cube is given as follows:
Surface area of a Cube = 6 side2
Given below are some of the examples that explains how to find surface area of a cube with the help of the surface area of a cube formula.
### Solved Example
Question: Find the surface area of a cube with side 4 inches.
Solution:
Side of a cube = 4
Surface area of a cube = 6 side2
Surface area of a cube = 6 * 42
= 6 * 4 * 4
= 96
Surface area of a cube = 96 sq. inches. |
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# 1.4 Fractions
So far we have thought of numbers in terms of their decimal form, e.g., 4.567, but this is not the only way of thinking of, or representing, numbers. A fraction represents a part of something. If you decide to share out something equally between two people, then each receives a half of the total and this is represented by the symbol ½.
A fraction is just the ratio of two numbers: 1/2, 3/5, 12/8, etc. We get the corresponding decimal form 0.5, 0.6, 1.5 respectively by performing division. The top half of a fraction is called the numerator and the bottom half the denominator, i.e., in 4/16, 4 is the numerator and 16 is the denominator. We divide the numerator (the top figure) by the denominator (the bottom figure) to get the decimal form. If, for instance, you use your calculator to divide 4 by 16 you will get 0.25.
A fraction can have many different representations. For example, 4/16, 2/8, and 1/4 all represent the same fraction, one quarter or 0.25. It is customary to write a fraction in the lowest possible terms. That is, to reduce the numerator and denominator as far as possible so that, for example, one quarter is shown as 1/4 rather than 2/8 or 4/16.
If we have a fraction such as 26/39 we need to recognise that the fraction can be reduced by dividing both the denominator and the numerator by the largest number that goes into both exactly. In 26/39 this number is 13 so (26/13) / (39/13) equates to 2/3.
We can perform the basic numerical operations on fractions directly. For example, if we wish to multiply 3/4 by 2/9 then what we are trying to do is to take 3/4 of 2/9, so we form the new fraction: 3/4 x 2/9 = (3 x 2) / (4 x 9) = 6/36 or 1/6 in its simplest form.
In general, we multiply two fractions by forming a new fraction where the new numerator is the result of multiplying together the two numerators, and the new denominator is the result of multiplying together the two denominators.
Addition of fractions is more complicated than multiplication. This can be seen if we try to calculate the sum of 3/5 plus 2/7. The first step is to represent each fraction as the ratio of a pair of numbers with the same denominator. For this example, we multiply the top and bottom of 3/5 by 7, and the top and bottom of 2/7 by 5. The fractions now look like 21/35 and 10/35 and both have the same denominator, which is 35. In this new form we just add the two numerators.
(3/5) + (2/7) = (21/35) + (10/35)
= (21 + 10) / 35
= 31/35
## Activity 6
### Part (a - i)
(a) Convert the following fractions to decimal form (rounded to three decimal places) by dividing the numerator by the denominator on your calculator:
(i) 125/1000
(i) 0.125
(ii) 8/24
(ii) 0.333
(iii) 32/36
(iii) 0.889
### Part (b - i)
(b) Perform the following operations between the fractions given:
(i) 1/2 x 2/3
(i) 1/3
### Part (b - ii)
(ii) 11/34 x 17/19
(ii) 187/646 = 11/38 (if top and bottom both divided by 17)
(iii) 2/5 x 7/11
(iii) 14/55
### Part (b - iv)
(iv) 1/2 + 2/3
(iv) 7/6 or 1 1/6
(v) 3/4 x 4/5 |
# Quadratic Techniques to Solve Polynomial Equations CCSS: F.IF.4 ; A.APR.3.
## Presentation on theme: "Quadratic Techniques to Solve Polynomial Equations CCSS: F.IF.4 ; A.APR.3."— Presentation transcript:
Quadratic Techniques to Solve Polynomial Equations CCSS: F.IF.4 ; A.APR.3
CCSS: F.IF.4 For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums.
CCSS: A.APR.3 Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial.
Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning.
Essential Question: How do we use quadratic techniques solving equations?
Objectives Solve third and fourth degree equations that contain quadratic factors, and Solve other non-quadratic equations that can be written in quadratic form.
Intro Some equations are not quadratic but can be written in a form that resembles a quadratic equation. For example, the equation x 4 – 20x 2 + 64 = 0 can be written as (x 2 ) 2 – 20x 2 + 64 = 0. Equations that can be written this way are said to be equations in quadratic form.
Key Concept: An expression that is quadratic form can be written as: au² +bu + c for any numbers a, b, and c, a≠0, where u is some expression in x. The expression au² +bu + c is called quadratic form of the original expression.
Once an equation is written in quadratic form, it can be solved by the methods you have already learned to use for solving quadratic equations.
Ex. 1: Solve x 4 – 13x 2 + 36 = 0 The solutions or roots are -3, 3, -2, and 2.
The graph of x 4 – 13x 2 + 36 = 0 looks like: The graph of y = x 4 – 13x 2 + 36 crosses the x-axis 4 times. There will be 4 real solutions.
Recall that (a m ) n = a mn for any positive number a and any rational numbers n and m. This property of exponents that you learned in chapter 5 is often used when solving equations.
Ex. 2: Solve
Ex. 3: Solve
Ex. 4: Solve
There is no real number x such that is = - 1.Since principal root of a number can not be negative, -1 is not a solution. The only solution would be 64.
Some cubic equations can be solved using the quadratic formula. First a binomial factor must be found.
Ex. 5: Solve
Quadratic form ax 2 + bx + c = 02x 2 – 3x – 5 = 0 This also would be a quadratic form of an equation
How would you solve Use Substitution
How would you solve Use Substitution Let u = y 2 u 2 = y 4
How would you solve Use Substitution Let u = y 2 u 2 = y 4
How would you solve Then Use Substitution Let u = y 2 u 2 = y 4
How would you solve Then Use Substitution Let Do both answers work?
How would you solve Then Use Substitution Let Do both answers work?
How would you solve Then Use Substitution Let
How would you solve Then Use Substitution Let Do both answers work? |
# Unit 6 Family Materials
Expressions and Equations
### Expressions and Equations
Here are the video lesson summaries for Grade 6, Unit 6: Expressions and Equations. Each video highlights key concepts and vocabulary that students learn across one or more lessons in the unit. The content of these video lesson summaries is based on the written Lesson Summaries found at the end of lessons in the curriculum. The goal of these videos is to support students in reviewing and checking their understanding of important concepts and vocabulary. Here are some possible ways families can use these videos:
• Keep informed on concepts and vocabulary students are learning about in class.
• Watch with their student and pause at key points to predict what comes next or think up other examples of vocabulary terms (the bolded words).
• Consider following the Connecting to Other Units links to review the math concepts that led up to this unit or to preview where the concepts in this unit lead to in future units.
Grade 6, Unit 6: Expressions and Equations
Vimeo
Video 1: Understanding Equations (Lessons 1–3)
Video 2: Writing and Solving Equations (Lessons 4–7)
Video 3: Writing Equivalent Expressions (Lessons 8–11)
Video 4: Expressions with Exponents (Lessons 12–15)
Video 5: Relationships Between Quantities (Lessons 16–18)
Video 1
Video 2
Video 3
Video 4
Video 5
### Equations in One Variable
This week your student will be learning to visualize, write, and solve equations. They did this work in previous grades with numbers. In grade 6, we often use a letter called a variable to represent a number whose value is unknown. Diagrams can help us make sense of how quantities are related. Here is an example of such a diagram:
Since 3 pieces are labeled with the same variable $$x$$, we know that each of the three pieces represent the same number. Some equations that match this diagram are $$x+x+x=15$$ and $$15=3x.$$
A solution to an equation is a number used in place of the variable that makes the equation true. In the previous example, the solution is 5. Think about substituting 5 for $$x$$ in either equation: $$5+5+5=15$$ and $$15=3 \boldcdot 5$$ are both true. We can tell that, for example, 4 is not a solution, because $$4+4+4$$ does not equal 15.
Solving an equation is a process for finding a solution. Your student will learn that an equation like $$15=3x$$ can be solved by dividing each side by 3. Notice that if you divide each side by 3, $$15 \div 3 = 3x \div 3$$, you are left with $$5=x$$, the solution to the equation.
Here is a task to try with your student:
Draw a diagram to represent each equation. Then, solve each equation.
$$\displaystyle 2y=11$$
$$\displaystyle 11=x+2$$
Solution:
$$\displaystyle y=5.5 \text{ or }y=\frac{11}{2}$$
$$\displaystyle x=9$$
### Equal and Equivalent
This week your student is writing mathematical expressions, especially expressions using the distributive property.
In this diagram, we can say one side length of the large rectangle is 3 units and the other is $$x + 2$$ units. So, the area of the large rectangle is $$3(x+2)$$. The large rectangle can be partitioned into two smaller rectangles, A and B, with no overlap. The area of A is 6 and the area of B is $$3x$$. So, the area of the large rectangle can also be written as $$3x + 6$$. In other words, $$\displaystyle 3(x+2)=3x+3\boldcdot2$$ This is an example of the distributive property.
Here is a task to try with your student:
Draw and label a partitioned rectangle to show that each of these equations is always true, no matter the value of the letters.
• $$5x+2x=(5+2)x$$
• $$3(a+b)=3a+3b$$
Solution:
Answers vary. Sample responses:
### Expressions with Exponents
This week your student will be working with exponents. When we write an expression like $$7^n$$, we call $$n$$ the exponent. In this example, 7 is called the base. The exponent tells you how many factors of the base to multiply. For example, $$7^4$$ is equal to $$7 \boldcdot 7 \boldcdot 7 \boldcdot 7$$. In grade 6, students write expressions with whole-number exponents and bases that are
• whole numbers like $$7^4$$
• fractions like $$\left(\frac17\right) ^ 4$$
• decimals like $$7.7^4$$
• variables like $$x^4$$
Here is a task to try with your student:
Remember that a solution to an equation is a number that makes the equation true. For example, a solution to $$x^5=30+x$$ is 2, since $$2^5=30+2$$. On the other hand, 1 is not a solution, since $$1^5$$ does not equal $$30+1$$. Find the solution to each equation from the list provided.
1. $$n^2=49$$
2. $$4^n=64$$
3. $$4^n=4$$
4. $$\left(\frac{3}{4}\right)^2=n$$
5. $$0.2^3=n$$
6. $$n^4=\frac{1}{16}$$
7. $$1^n=1$$
8. $${3^n} \div {3^2}=3^3$$
List: $$0, 0.008, \frac12, \frac{9}{16}, \frac{6}{8}, 0.8, 1, 2, 3, 4, 5, 6, 7$$
Solution:
1. 7, because $$7^2=49$$. (Note that -7 is also a solution, but in grade 6 students aren’t expected to know about multiplying negative numbers.)
2. 3, because $$4^3=64$$
3. 1, because $$4^1=4$$
4. $$\frac{9}{16}$$, because $$\left(\frac{3}{4}\right)^2$$ means $$\left(\frac{3}{4}\right) \boldcdot \left(\frac{3}{4}\right)$$
5. 0.008, because $$0.2^3$$ means $$(0.2) \boldcdot (0.2) \boldcdot (0.2)$$
6. $$\frac12$$, because $$\left(\frac{1}{2}\right)^4=\frac{1}{16}$$
7. Any number! $$1^n=1$$ is true no matter what number you use in place of $$n$$.
8. 5, because this can be rewritten $${3^n} \div 9=27$$. What would we have to divide by 9 to get 27? 243, because $$27 \boldcdot 9=243$$. $$3^5=243$$.
### Relationships Between Quantities
This week your student will study relationships between two quantities. For example, since a quarter is worth 25ȼ, we can represent the relationship between the number of quarters, $$n$$, and their value $$v$$ in cents like this:
$$\displaystyle v = 25n$$
We can also use a table to represent the situation:
$$n$$ $$v$$
1 25
2 50
3 75
Or we can draw a graph to represent the relationship between the two quantities:
Here is a task to try with your student:
A shopper is buying granola bars. The cost of each granola bar is $0.75. 1. Write an equation that shows the cost of the granola bars, $$c$$, in terms of the number of bars purchased, $$n$$. 2. Create a graph representing associated values of $$c$$ and $$n$$. 3. What are the coordinates of some points on your graph? What do they represent? Solution: 1. $$c=0.75n$$. Every granola bar costs$0.75 and the shopper is buying $$n$$ of them, so the cost is $$0.75n$$.
2. Answers vary. One way to create a graph is to label the horizontal axis with "number of bars" with intervals, 0, 1, 2, 3, etc, and label the vertical axis with "total cost in dollars" with intervals 0, 0.25, 0.50, 0.75, etc.
3. If the graph is created as described in this solution, the first coordinate is the number of granola bars and the second is the cost in dollars for that number of granola bars. Some points on such a graph are $$(2,1.50)$$ and $$(10,7.50)$$ |
Harder Example – Alright, so now it’s time to look at an example where we are asked to find both the average rate of change and the instantaneous rate of change. Average And Instantaneous Rate Of Change Of A Function – Example Notice that for part (a), we used the slope formula to find the average rate of change over the interval. In contrast, for part (b), we used the power rule to find the derivative and substituted the desired x-value into the derivative to find the instantaneous rate of change. Nothing to it!
### What is the formula for average rate of change?
The average rate of change represents a measurement that can provide insight into a variety of applications. From finance and accounting to engineering applications, you can calculate the average rate of change using the simple algebraic formula: (y1 – y2) / (x1 – x2).
#### What is the average rate of change?
What is average rate of change? It is a measure of how much the function changed per unit, on average, over that interval. It is derived from the slope of the straight line connecting the interval’s endpoints on the function’s graph.
## How do you find the average rate of change from a table?
To find the average rate of change from a table or a graph we first identify the given intervals, find the change in the function’s y-values, the change in x-values, and, finally divide those find out rate.
#### How do you find the average rate of change between two points?
The average rate of change between two points is calculated as the slope of the straight line which connects the two points. To find the average rate of change of between and, use the formula f ( b ) − f ( a ) b − a.
#### What is an example of the average rate of change?
What is the Average Rate of Change? – The average rate of change of a function f(x) over an interval is defined as the ratio of “change in the function values” to the “change in the endpoints of the interval”.i.e., the average rate of change can be calculated using / (b – a).
In other words, the average rate of change (which is denoted by A(x)) is the “ratio of change in outputs to change in inputs”.i.e., A(x) = (change in outputs) / (change in inputs) = Δy / Δx = / (b – a) Here, Δy is the change in y-values (or) change in the function values and Δx is the change in x-values (or) the change in the endpoints of the interval.
Some examples of the average rate of change are:
A bus travels at a speed of 80 km per hour. The number of fish in a lake increases at the rate of 100 per week. The current in an electrical circuit decreases 0.2 amperes for a decrease of 1-volt voltage.
## What is an example of a rate of change?
Other examples of rates of change include: A population of rats increasing by 40 rats per week. A car traveling 68 miles per hour (distance traveled changes by 68 miles each hour as time passes) A car driving 27 miles per gallon of gasoline (distance traveled changes by 27 miles for each gallon)
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## Is slope the same as rate of change?
When finding the slope of real-world situations, it is often referred to as rate of change. ‘Rate of change’ means the same as ‘slope.’ If you are asked to find the rate of change, use the slope formula or make a slope triangle.
### How do you calculate average change in Excel?
Want more? – Calculate the average of a group of numbers AVERAGE function AVERAGEIF function You may have used AutoSum to quickly add numbers in Excel. But did you know you can also use it to calculate other results, such as averages? Click the cell to the right of a row or below a column.
Then, on the HOME tab, click the AutoSum down arrow, click Average, verify the formula if what you want, and press Enter. When I double-click inside the cell, I see it is a formula with the AVERAGE function. The formula is AVERAGE, A2, colon, A5, which averages the cells from A2 through A5. When averaging a few cells, the AVERAGE function saves your time.
With larger ranges of cells, it’s essential. If I try to use the AutoSum, Average option here, it only uses cell C5, not the entire column. Why? Because C4 is blank. If C4 contained a number, C2 through C5 would be an adjacent range of cells that AutoSum would recognize.
To average values in cells and ranges of cells that aren’t adjacent, type an equals sign (a formula always starts with an equals sign), AVERAGE, open parenthesis, hold down the Ctrl key, click the desired cells and ranges of cells, and press Enter. The formula uses the AVERAGE function to average the cells containing numbers and ignores the empty cells or those containing text.
For more information about the AVERAGE function, see the course summary at the end of the course. Up next, AVERAGEIF,
### How do you find the average rate of change without a graph?
Harder Example – Alright, so now it’s time to look at an example where we are asked to find both the average rate of change and the instantaneous rate of change. Average And Instantaneous Rate Of Change Of A Function – Example Notice that for part (a), we used the slope formula to find the average rate of change over the interval. In contrast, for part (b), we used the power rule to find the derivative and substituted the desired x-value into the derivative to find the instantaneous rate of change. Nothing to it!
## How do you find the average of a table?
Mean from a frequency table GCSE questions – 1. Alex works in a restaurant. At 1 pm one day she records the number of customers sitting at the tables in the restaurant. Here are the results. (a) Work out the total number of tables in the restaurant. (b) Work out the total number of customers sitting at the tables in the restaurant. (c) Work out the mean number of customers per table in the restaurant. (5 marks) Show answer (a) 25 25 For the correct answer (1) (b) ( 0 × 3 ) + ( 1 × 4 ) + ( 2 × 10 ) + ( 3 × 3 ) + ( 4 × 5 ) (0\times 3)+(1\times 4)+(2\times 10)+(3\times 3)+(4\times 5) For the sum of the products (1) = 53 =53 For the correct answer (1) (c) = 53 ÷ 25 =53\div 25 For the division (1) = 2.12 =2.12 For the correct answer (1) 2. Work out an estimate for the mean number of minutes late. (3 marks) Show answer 5, 15 5, 15 and 25 25 For the midpoints (1) ( 5 × 5 ) + ( 3 × 15 ) + ( 2 × 25 ) 10 \frac For the sum of the products divided by 10 (1) = 120 10 = 12 =\frac =12 For the correct answer (1) 3. The table shows the weight, w w kg of 90 90 bags that people took on a coach trip. Calculate an estimate for the mean weight of the 90 90 bags. Give your answer to 3 3 significant figures. (4 marks) Show answer 2.5, 7.5, 12.5, 17.5 2.5, 7.5, 12.5, 17.5 and 22.5 22.5 For the midpoints (1) ( 12 × 2.5 ) + ( 19 × 7.5 ) + ( 23 × 12.5 ) + ( 31 × 17.5 ) + ( 5 × 22.5 ) (12\times 2.5)+(19\times 7.5)+(23\times 12.5)+(31\times 17.5)+(5\times 22.5) For the sum of the products (1) = 1115 90 = 12.388 =\frac =12.388 For the dividing the sum of the products by 90 90 (1) = 12.388 = 12.4 (3 s.f.) =12.388=12.4 \ \text For the correct answer (1)
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## How do you find the rate of change between two numbers?
To calculate percentage decrease: – First: work out the difference (decrease) between the two numbers you are comparing. Decrease = Original Number – New Number Then: divide the decrease by the original number and multiply the answer by 100. % Decrease = Decrease ÷ Original Number × 100 If your answer is a negative number, then this is a percentage increase.
#### How do you find the average of two rates?
To find the average percentage of the two percentages in this example, you need to first divide the sum of the two percentage numbers by the sum of the two sample sizes. So, 95 divided by 350 equals 0.27. You then multiply this decimal by 100 to get the average percentage.
### What is the formula in finding the average?
Remarks –
Arguments can either be numbers or names, ranges, or cell references that contain numbers. Logical values and text representations of numbers that you type directly into the list of arguments are not counted. If a range or cell reference argument contains text, logical values, or empty cells, those values are ignored; however, cells with the value zero are included. Arguments that are error values or text that cannot be translated into numbers cause errors. If you want to include logical values and text representations of numbers in a reference as part of the calculation, use the AVERAGEA function. If you want to calculate the average of only the values that meet certain criteria, use the AVERAGEIF function or the AVERAGEIFS function.
Note: The AVERAGE function measures central tendency, which is the location of the center of a group of numbers in a statistical distribution. The three most common measures of central tendency are:
Average, which is the arithmetic mean, and is calculated by adding a group of numbers and then dividing by the count of those numbers. For example, the average of 2, 3, 3, 5, 7, and 10 is 30 divided by 6, which is 5. Median, which is the middle number of a group of numbers; that is, half the numbers have values that are greater than the median, and half the numbers have values that are less than the median. For example, the median of 2, 3, 3, 5, 7, and 10 is 4. Mode, which is the most frequently occurring number in a group of numbers. For example, the mode of 2, 3, 3, 5, 7, and 10 is 3.
For a symmetrical distribution of a group of numbers, these three measures of central tendency are all the same. For a skewed distribution of a group of numbers, they can be different. Tip: When you average cells, keep in mind the difference between empty cells and those containing the value zero, especially if you have cleared the Show a zero in cells that have a zero value check box in the Excel Options dialog box in the Excel desktop application.
On the File tab, click Options, and then, in the Advanced category, look under Display options for this worksheet,
## Is average rate of change a percentage?
Percentage change – When you have data for two points in time, you can calculate how much change there has been during this period. The result is expressed as a percentage (in absolute numbers, it’s just a difference) and is called the rate of change, i.e. the percentage change, It is calculated as follows: × 100.
#### What are the two types of rate of change?
Rate of Change A rate of change is a rate that describes how one quantity changes in relation to another quantity. If x is the independent variable and y is the dependent variable, then rate of change = change in y change in x Rates of change can be positive or negative.
## What are 3 examples of a rate?
Here are some examples of common rates: Speed limit, interest rate, crime rate, profit rate, birth rate, death rate, etc. Maybe there’s a unit of time or if you’re looking at heart rate (beats per minute). Rates are everywhere in our lives, which is why they are so important to understand.
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#### What is the rate of change of a function?
The rate of change function is defined as the rate at which one quantity is changing with respect to another quantity. In simple terms, in the rate of change, the amount of change in one item is divided by the corresponding amount of change in another.
#### How do you calculate average change in Excel?
Want more? – Calculate the average of a group of numbers AVERAGE function AVERAGEIF function You may have used AutoSum to quickly add numbers in Excel. But did you know you can also use it to calculate other results, such as averages? Click the cell to the right of a row or below a column.
• Then, on the HOME tab, click the AutoSum down arrow, click Average, verify the formula if what you want, and press Enter.
• When I double-click inside the cell, I see it is a formula with the AVERAGE function.
• The formula is AVERAGE, A2, colon, A5, which averages the cells from A2 through A5.
• When averaging a few cells, the AVERAGE function saves your time.
With larger ranges of cells, it’s essential. If I try to use the AutoSum, Average option here, it only uses cell C5, not the entire column. Why? Because C4 is blank. If C4 contained a number, C2 through C5 would be an adjacent range of cells that AutoSum would recognize.
• To average values in cells and ranges of cells that aren’t adjacent, type an equals sign (a formula always starts with an equals sign), AVERAGE, open parenthesis, hold down the Ctrl key, click the desired cells and ranges of cells, and press Enter.
• The formula uses the AVERAGE function to average the cells containing numbers and ignores the empty cells or those containing text.
For more information about the AVERAGE function, see the course summary at the end of the course. Up next, AVERAGEIF,
#### What is the formula in finding the average?
Remarks –
Arguments can either be numbers or names, ranges, or cell references that contain numbers. Logical values and text representations of numbers that you type directly into the list of arguments are not counted. If a range or cell reference argument contains text, logical values, or empty cells, those values are ignored; however, cells with the value zero are included. Arguments that are error values or text that cannot be translated into numbers cause errors. If you want to include logical values and text representations of numbers in a reference as part of the calculation, use the AVERAGEA function. If you want to calculate the average of only the values that meet certain criteria, use the AVERAGEIF function or the AVERAGEIFS function.
Note: The AVERAGE function measures central tendency, which is the location of the center of a group of numbers in a statistical distribution. The three most common measures of central tendency are:
Average, which is the arithmetic mean, and is calculated by adding a group of numbers and then dividing by the count of those numbers. For example, the average of 2, 3, 3, 5, 7, and 10 is 30 divided by 6, which is 5. Median, which is the middle number of a group of numbers; that is, half the numbers have values that are greater than the median, and half the numbers have values that are less than the median. For example, the median of 2, 3, 3, 5, 7, and 10 is 4. Mode, which is the most frequently occurring number in a group of numbers. For example, the mode of 2, 3, 3, 5, 7, and 10 is 3.
For a symmetrical distribution of a group of numbers, these three measures of central tendency are all the same. For a skewed distribution of a group of numbers, they can be different. Tip: When you average cells, keep in mind the difference between empty cells and those containing the value zero, especially if you have cleared the Show a zero in cells that have a zero value check box in the Excel Options dialog box in the Excel desktop application.
On the File tab, click Options, and then, in the Advanced category, look under Display options for this worksheet,
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# Example of greater than and less than: General Data Protection Regulation(GDPR) Guidelines BYJU’S
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## Greater Than and Less Than Symbols For Preschoolers and Kids
• What Is Greater Than Sign (>)?
• What Is Less than Sign (<)?
• What Is Equal To Sign (=)?
• Rules For Comparison Of Numbers
• Why Do We Use The Greater Than And Less Than?
• Examples Of Greater Than And Less Than
• Simple Tricks To Remember Greater And Less Than Signs
• Word Problems On Greater Than & Less Than Symbols For Kids
• Practice Questions To Help Kids Revise Greater Than & Less Than Sign
• Activities To Teach Your Child Greater Than and Less Than Symbols
Mathematics is a language with its own sets of rules and formulas. The symbols and signs used in maths are quite unique and are universally accepted. From school till graduation, we encounter these signs in all the mathematical concepts we study. The use of these signs and symbols helps us to say what we want, consuming less of our time. For example, greater than and less than signs help us convey the meaning using the signs “>” and “<“.
In this article, we are going to learn about greater than and less than concept with the help of their definition, their signs, solved examples and some fun activities. The greater and less concept is used for comparing two numbers. Therefore, it is important that children understand how to use and represent greater than and less than using signs.
Let’s get started and teach greater than and less than to kindergarten kids.
### What Is Greater Than Sign (>)?
Greater than sign helps us to define which quantity/amount is larger than the other one. The greater than sign in maths is “>”. This sign is placed between two values in which one number is greater than the other number.
### What Is Less than Sign (<)?
Less than sign helps us to define which quantity/amount is smaller than the other one. The less than sign in maths is “< “. This sign is placed between two values in which one number is smaller than the other number.
### What Is Equal To Sign (=)?
Equal to sign helps us to define the equality between two numbers or values. The equal to sign in maths is “=”. The sign is placed between two values when they are equal.
### Rules For Comparison Of Numbers
Here are some rules on how to read greater than and less than numbers.
• The more the number of digits, the greater the number.
• If two numbers have an equal number of digits, look for the left-hand side of the two numbers. The number with the bigger digit on the left-hand side is greater.
• If the two leftmost digits are the same, then compare the next digit to the right and continue to compare the numbers in this manner.
### Why Do We Use The Greater Than And Less Than?
Greater than and less than are used to compare any two numbers/values/amounts/etc. When the given number is bigger or smaller than the other number, greater than and less than signs are used. If the first number is greater than the second number, greater than sign (>) is used. If the first number is less than the second number, less than the sign (<) is used.
### Examples Of Greater Than And Less Than
It is important that we teach our little ones how to use the greater than and less than signs. Teaching and learning with the help of examples make the learning process easy for children. In this section, we have covered examples of greater than and less than for children. Let’s read and teach!
#### Greater Than Symbol Examples
• 5 > 3 (5 is greater than 3)
• 7 > 5 (7 is greater than 5)
• 2 > 1 (2 is greater than 1)
• 20 > 15 (20 is greater than 15)
• 60 > 35 (60 is greater than 35)
• 100 > 70 (100 is greater than 70)
• 22 > 13 (22 is greater than 13)
• 9 > 3 (9 is greater than 3)
#### Less Than Symbol Examples
• 2 < 4 (2 is less than 4)
• 17 < 34 (17 is less than 34)
• 23 < 74 (23 is less than 74)
• 89 < 100 (89 is less than 100)
• 68 < 72 (68 is less than 72)
• 24 < 42 (24 is less than 42)
• 99 < 100 (99 is less than 100)
• 25 < 40 (25 is less than 40)
### Simple Tricks To Remember Greater And Less Than Signs
Let’s read about some tricks to remember greater and less than signs for kids. Here we will discuss 3 methods:
#### 1. Understanding signs using points
For Greater Than Sign (>)
• Understand the sign from the left side to the right side.
• On the left hand, it has two points, and on the right side, it has one point, which means it denotes greater than the sign.
For Less Than Sign (<)
• Understand the symbol from the left side to the right side.
• On the left hand, it has one point, and on the right side, it has two points, which means it denotes less than a sign.
#### 2. Alligator Method
• We assume the sign “<” and “>” are alligators with a number on both sides, which are fishes.
• The alligator always wants to eat a larger number of fish, so the mouth is always opens towards the greater number.
#### 3. L Method
• The letter ‘L’ looks like the less-than symbol “<“.
• You can remember the first letter of the word, less than to be the sign. Example: 10 < 50 (10 is less than 50)
Sign Meaning Sign Example Less than sign < 10 < 20 More than sign > 20 > 10 Equal to sign = 20 = 20
### Word Problems On Greater Than & Less Than Symbols For Kids
Here are some word problems with the solution on greater than and less than symbols for children.
1. Ram has twelve bananas, and Mansi has six bananas. Find out who has fewer bananas.
Solution:
Ram has 12 bananas.
Mansi has 6 bananas.
So, 6 is less than 12, 6 <12
Therefore, Mansi has less bananas than Ram.
2. Gayatri walks in the park for fifty minutes, while Rohan walks in the park for 30 minutes every day. Who walks for less time in the park?
Solution:
Gayatri walks for 50 minutes.
Rohan walks for 30 minutes.
Therefore, Rohan walks for less time in the park.
So, 30 is less than 50, 30 < 40.
3. Sita purchased two bags, while Geeta purchased only one bag. Who purchased less bags?
Solution:
Sita purchased 2 bags.
Geeta purchased 1 bag.
So, 1 is less than 2, 1 < 2.
Therefore, Geeta purchased less bags.
4. Giraffe’s weight is 1800 kg, and a zebra’s weight is 300 kg. Who weighs more?
Solution:
Giraffe weighs 1800 kg.
Zebra weighs 300 kg.
So, 1800 is greater than 300, 1800 > 300
So, giraffe weight is more than that of zebra.
### Practice Questions To Help Kids Revise Greater Than & Less Than Sign
1. Place the correct sign:
20 _ 30
50 _ 25
80 _ 80
20 < 30
50 > 25
80 = 80
2. Rani has 5 apples; Jahnvi has 9 apples. Fill in the correct sign to tell who has more apples.
9 _ 5
9 > 5
Therefore, Jahnvi has a more number of apples than Rani.
3. 77 _ 77. Fill in the correct sign.
4. Use the correct sign and tell which number is smaller.
100 _ 250
100 < 250
Therefore, 100 is less than 250.
5. Between the numbers 44 and 66, which is greater?
Therefore, 66 is greater than 44.
6. Fill in the correct symbol.
90 _ 82
90 > 82
### Activities To Teach Your Child Greater Than and Less Than Symbols
Let’s take a look at some greater than less than preschool activities.
#### 1. Throw The Dice
Take two dice and ask your child to throw them to draw two numbers. Ask them which number is greater or less than the other number.
#### 2. Count And Tell
Collect your little one’s soft toys and storybooks. Keep the two things in separate stacks. Ask your child to count each of the two categories and tell which one is less in number than the other one.
#### 3. Activity Worksheets
Make an activity worksheet for your child where they have to just put in the sign of greater than and less than. For example, you can make a similar activity sheet for your child:
• 9 _ 8
• 55 _ 89
• 21 _ 12
• 77 _ 77
• 99 _ 100
• 15 _ 62
#### 4. Card Game
Start playing this card game with your child. Divide the cards equally between you and your child. Both of you have to throw one card at a time. Ask your child to compare the number on the two cards and tell which one is greater than the other.
#### 5. Sensory Bin Play
Take a basket and add pom poms of any 2 colours to it. Ask your child to sort the pom poms according to their colours and tell which one coloured pom poms are less than the other one.
We hope this article on greater than and less than (more than and less than) helped your little one to learn and understand the concept easily. The activities mentioned above are very simple, fun and engaging and will help your child in learning an important mathematical concept. If you have any questions related to this topic, drop them in the comments below, and we will be happy to provide you with answers.
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## Basic Math Symbols Explained | HowStuffWorks
«»
Common mathematical symbols are the building blocks of all mathematical functions. Bankrx/Shutterstock
Common math symbols give us a language for understanding, well, everything from budgeting to the nature of reality itself. Its building blocks are relatively simple. Even the most sophisticated mathematical equations rely on a handful of fundamental common math symbols.
Before you can solve the mystery of the Collatz Conjecture, figure out a square root or understand more complex algebraic symbols, you’ll need to master the basic mathematical symbols that are necessary for writing a mathematical equation.
Contents
1. Plus Symbol (+)
2. Minus Symbol (-)
3. Equals Symbol (=)
4. Does Not Equal Symbol (≠)
5. Multiplication Symbol (×)
6. Division Symbol (÷)
7. Greater Than/Less Than Symbols
8. Greater Than or Equal To/Less Than or Equal To Symbols (≥ ≤)
9. Fraction Symbol (/)
10. Decimal Symbol (. )
11. Percent Symbol (%)
### 1. Plus Symbol (+)
The plus symbol (+) signifies addition. It’s the most basic math symbol in the world. When adding two or more numbers, use the plus symbol to indicate that you are combining them. For example, 6 + 3 means you’re adding positive number 6 and positive number 3 together. You can also include the plus symbol before a number to indicate that the number is positive, although this is typically redundant — a number on its own is assumed to be positive. Writing «+3» is nonetheless a way of making clear that you’re referring to positive 3.
### 2. Minus Symbol (-)
The minus symbol (-) signifies subtraction. When you’re subtracting one number from another, place the minus sign between them. For instance, 6 — 3 shows that you’re subtracting 3 from 6. As with the plus symbol, you can place the minus symbol in front of a number to show that it has a negative value. This is much more common, since written numbers are not negative by default. As an example, writing «-3» shows that you’re referring to negative 3.
### 3. Equals Symbol (=)
The equals symbol (=) indicates that the values on either side of the symbol are not approximately equal, but are completely equivalent. In the equation 6 + 3 = 9, the equals sign indicates that the sum of 6 and 3 is equivalent to 9. The equals symbol is an essential part of any math equation.
### 4.
Does Not Equal Symbol (≠)
The does not equal symbol (≠) indicates that two values are not equal. Place this sign between two numbers or mathematical expressions that are not equivalent. For example, 6 ≠ 3 states that 6 is not equal to 3.
### 5. Multiplication Symbol (×)
The multiplication symbol (×) signifies multiplying something by something else — that is, finding the product of two numbers or, to put it another way, adding a number to itself a certain number of times. Let’s make that clear with an example: 6 × 3 = 18 means you’re adding three 6s together, resulting in a product of 18. Since the formal multiplication symbol (×) is not common on keyboards, you can use an asterisk (*) or an «x» instead. This is particularly useful when writing computer programs or Excel formulas.
### htm»> 6. Division Symbol (÷)
The division symbol (÷) signifies the dividing of a number. This is the process of splitting a number into a certain number of equal parts. Consider the equation 6 ÷ 3 = 2. In this example, 6 divides into 3 equal groups of 2. Like one of the other key mathematical objects, the multiplication symbol, the formal symbol for division (÷) is uncommon in everyday use. When typing out equations, you can use a forward slash (/) to indicate division. Again, this is necessary for writing equations in computer programming languages.
«»
Math equations can often look like a confusing mass of random squiggles, but they are all made up of common mathematical symbols.
Margarita Vin/Shutterstock
### 7. Greater Than/Less Than Symbols
The greater than symbol (>) and the less than symbol (<) don’t have the same meaning, but indicate that one value is greater than another. These symbols function similarly to an equals symbol in between two numbers. For example, 6 > 3 shows that 6 is greater than 3, while 3 < 6 shows that 3 is less than 6. Remember, the larger number always faces the open end of the symbol, while the smaller number always faces the point where the two lines meet.
### 8. Greater Than or Equal To/Less Than or Equal To Symbols (≥ ≤)
The greater than or equal to symbol (≥) and the less than or equal to symbol (≤) combine the greater than and less than symbols with the equals symbol. They’re used to, you guessed it, show when two values are greater (or less) than or equal to each other. This symbol is not very common in everyday usage, and it’s most prevalent in equations when one or more quantities is unknown. For example, in the equation X ≥ 3, we know that X can be 3 or any number greater than 3. In this case, 3 ≥ 3 is a true statement, as is 4 ≥ 3, as is 5 ≥ 3, and so on.
### 9. Fraction Symbol (/)
The fraction symbol (/) appears as a line or slash separating two numbers, one below the other. It can appear in a few different ways. For instance, 3/5 means three-fifths. The 3 at the top of the fraction is in the position of the numerator, and the five at the bottom of the fraction is in the position of the denominator. Fractions show you how many parts of a whole you have; saying that you have 3/5 of a cookie means that if a cookie is divided into five equal parts, you have 3 of those parts. For more complicated math expressions, the fraction symbol appears as a long horizontal line separating the numerator and denominator.
### htm»> 10. Decimal Symbol (.)
A decimal (.) symbol is a period symbol used to separate the whole part of a number from the fractional part of a number. If that sounds a bit confusing, let’s take a step back to make sense of it. The number system is based on a system of place value, meaning that the placement of each digit within a number indicates its value. In the number 3.6, the placement of the 3 indicates that is the whole part of the number; the 6 is to the right of the decimal in what we call the «tenths place,» meaning it is 6/10 of 1. If you had 3.6 cookies, you would have 3 and 6/10 total cookies. Additional digits after the decimal have their own place value. In the number 3.687, 8 is in the hundredths place, and 7 is in the thousandths place.
### 11. Percent Symbol (%)
Like the fraction symbol and decimal, the percent symbol (%) is one of the key mathematical objects, useful for showing fractional quantities, in this case specifically as a portion of 100. If you have 36% of your cell phone battery, you have 36 out of 100 units of battery life remaining. «Percent» means «out of one hundred,» and since the percent symbol (%) looks like the digits of 100 rearranged, it’s easy to remember.
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Thomas Harlander
«11 Basic Math Symbols and How to Use Them»
14 April 2023.
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29 June 2023
Citation
## Mathematics Signs greater than, less than and equal to
### Lesson outline
Topic: Signs greater than, less than and equal to
9000 8 Lesson progress
1. Organizational start of the lesson
To study well,
There is a lot to know.
Every day piggy bank knowledge
Definitely replenish.
2. Knowledge update Let’s learn how to compare things. And a fairy tale hero will help us with this. Guess which one.
As mischievous as the letter «B»,
The passenger is a little strange,
Wooden man,
On land and underwater
Looking for a golden key.
He has a long nose everywhere.
Who is this?..
. Pinocchio!
1) Mental count
— Count from 10 to 30 and vice versa.
What numbers come after three, four, two? Four, five, three .
What numbers come before the numbers one, five, two? Zero, four, one .
What number are neighbors of numbers 4 and 6? Five .
— Two and four? Three .
— Zero and two? One .
— Three and five? Four .
— Solve the examples.
— One plus two equals — … three .
— Two plus two equals — … four .
— Three add two — … five .
— Four add one — … five .
— From five, subtracting two will be — … three .
— Four subtract three equals — … one .
— Three subtract one equals — … two .
— Five subtract three equals — … two .
— Four minus two will be — … two .
— Help Pinocchio cope with examples.
4 — 1 = 3
5 — 2 = 3
3 + 1 = 4
3 + 2 = 5
2 + 2 \u003d 4
— Look carefully at the pictures and select examples for them.
I draw a cat house:
Three windows, a door with a porch.
Above is another window,
To not be dark.
Count windows
In the cat’s house.
— Four, since adding one to three makes four.
A gray cat is sitting by the window,
A red cat lies on a matting.
A black cat plays with a mouse,
A white cat climbed into the basket.
How many cats?
Don’t hesitate, look.
How much did you count?
Speak!
— Four, because one plus one, plus one, plus one makes four.
— Two girls and three boys were riding the hill. How many children were on the slide?
Two and even three will be — five.
— The two boys are gone. How many children are left on the hill?
Five without two will be three.
— How many squares are drawn?
Of course, five.
Physical education for fingers
Here are all my fingers,
Turn them however you want.
And like this, and like this,
Don’t get offended at all.
One, two, three, four, five —
They don’t fit again.
Knocked,
Turned
And they wanted to work.
Let the handles rest,
And now back on the road.
2. Formation of knowledge
— How many mushrooms are on the left? Three .
— How many mushrooms are on the right? Two .
— Where are the most mushrooms? Of course, on the left.
— We can indicate this with the sign « is greater than «. This sign opens its «beak» towards a larger number.
— We read like this: «Three is more than two.» Write it down by saying it again.
What should we do when the number is
More like out of spite?
How to show it,
For everyone to understand?
Here for this, friends,
«More» sign I draw.
He is from a larger number
Shoots like an arrow,
And points us to
For the one who is smaller.
— How many apples are on the left? Four .
— How many apples are on the right? Five .
— Where are the least apples? Left .
— We can write it down with the sign «9»0009 is less than
.» This sign is turned with a corner towards a smaller number, that is, a closed “beak” indicates a smaller number. We read like this: «Four is less than five.» Write it down by saying it again.
Lucky for the big number.
Where is the smaller number?
And everyone knows him,
«Less than» sign denotes.
This is the same sign,
But it costs quite differently:
As if doing a somersault,
Rolls the number to the left side.
This means she is
It just has to be smaller.
— How to make sure that there are the same number of mushrooms?
That’s right, you need to add or remove one mushroom.
— This can be written using the sign « equals «. We read like this: “Three is equal to three”
— How to level apples?
— Read the inscription and write it down.
If we simply compare
Two numbers, one with the other,
And we will see that they are
Equal in value, —
We put it like that,
Between them is an «equal» sign.
This sign, remember you,
Looks like two dashes.
He has such fame:
Left as much as right.
— How many balls are left? Two .
— How many balls are on the right? One .
Which side has more balls? More balls on the left.
— Why is the number «one» on the right? Correct, because there is one ball on the right.
Is the «greater than» sign correct? Prove it.
Of course, right. After all, two is more than one, so the “beak” is open towards a larger number.
Physical education minute
So many times we chop wood.
We squat so many times,
How many balls do we have.
How many circles will I show,
So many jumps.
3. Consolidation of knowledge
— Which is less: cars or trucks? That’s right, trucks.
— How to write it down? Three is less than four.
— What more: cars or buses? That’s right, cars.
— And how to write it down? Four is more than one.
— If two more buses arrived, why are there more buses or cars?
That’s right, there are more cars. This can be written as follows: one plus two is less than four.
— All these entries are called inequalities .
— What is more now: buses or trucks? They are the same number.
— These entries are called equals .
Reference
4 — 1 = 3 3 + 1 1
5 — 1 > 2 4 + 1 = 5 1 + 2 = 3
— Write equals in one column and inequalities in the other.
— Check yourself!
4 — 1 = 3 5 — 1 > 2
4 + 1 = 5 3 + 1
1 + 2 = 3 1 + 2 > 1
Reference
1 2 3 1 1 3 2 1
— Sort the numbers by floor.
— Check yourself!
1 3 2 2 3 1 1 4 2 3 3 2 4 1
— How many sticks do you need to take to build a triangle? Three .
— Take four counting sticks and build a new figure.
What is this figure called?
It is called quadrilateral because it has four sides and four vertices. If a quadrilateral has right angles, then it is called rectangle .
— If I take six sticks and lay out such a figure, what will it be called? Why?
That’s right — it’s a hexagon, because it has six sides and six vertices.
Reference
▲▲ ■■■■ ○○○○○ ▲▲ □□□
▲ □□ ○○○ □□□□
2 > 1 4 … 2 + 2 5 … 4 — 1 3 … 4
— Compare the objects, write down the inequalities and insert the necessary signs.
— Check yourself!
— Yura and Olya measured the distance from the house to the tree using steps. Why did they get different answers?
That’s right, they have different stride lengths.
Reference
1 + 3 5 — 4
2 + 1 4 — 2
4 + 1 3 — 2
1 + 1 5 — 1
— Hang the buckets on the yokes, having previously solved the example.
— Check yourself!
— One donkey carried 10 kilograms of sugar, and the other — 10 kilograms of cotton wool. Who had the heaviest load?
The load was the same. Despite the fact that the sizes are different, but the weight is the same.
4. Summing up
— In which hand is the sign “In the left.
— In which hand is the «>» sign? On the right.
What does the «=» sign mean? Equality.
3 + 2 = 5 4 > 3
4 = 4 1
— Name the equalities.
Three plus two equals five. Four equals four.
— Name the figures.
Reflection
Today at the lesson:
I found out…
I learned …
I was wondering…
It was difficult for me…
Pinocchio says goodbye to you. See you soon!
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## Table of mathematical signs (symbols): values, print
MicroExcel. ru Mathematics Basic Mathematical Signs and Symbols
Below is a table with the main mathematical symbols and signs: root (√), greater than (>), less than (<), equality (=), etc. Also, their brief description and examples are given for better understanding. 9 Name Meaning / description
9 0717 Example = equals equals е. 4 плюс 3 равно 7</nobr>» data-order=»<nobr>7 = 4 + 3,</nobr><br><nobr>т.е. 4 плюс 3 равно 7</nobr>»> 7 = 4 + 3,
t. e. 4 plus 3 equals 7 ≠ not equal to inequality 7 ≠ 10,
7 not equal to 10 9074 3 ≈ approximately equals approximate equality/rounding равно 0,36</nobr>» data-order=»<nobr>0,35765 ≈ 0,36,</nobr><br><nobr>0,35765 прибл. равно 0,36</nobr>»> 0.35765 ≈ 0.36,
0.35765 approx. equal to 0.36 > greater than A greater than B 15 > 10,
15 greater than 10 < 90 397 less than A less than B 6 < 8,
6 less than 8 ≥ greater than or equal to A greater than or equal to B 10 ≥ 4,
10 greater than or equal to 4 ≤ less than or equal to A less than or equal to B 3 ≤ 7,
3 less than or equal to 7
parentheses expression inside parentheses is considered first 3 ⋅ (4 + 6) = 30 [ ] square brackets the expression inside the brackets is considered first [(1 + 3) ⋅ (2 + 4)] = 24
9039 3
+ plus sign of addition 1 + 2 = 3 − minus sign of subtraction 3 − 2 = 1 9 0838 ± plus or minus Both addition and subtraction are performed 4 ± 6 = 10 or -2 ± minus-plus both subtraction and addition are performed 5 ∓ 8 = -3 or 13 * sprocket x 3 * 3 = 9 x x x 3 x 3 = 9
9039 3
⋅ point multiplication 3 ⋅ 3 = 9 ÷ obelus division 8 ÷ 2 = 4 / slash division 8 / 2 = 4 : colon division 8 : 2 = 4 — horizontal line fraction (division) 1/2 mod 9 0397
modulo modulo 7 mod 2 = 1,
7 : 2 = 3 (remainder 1) » data-order=».»> . point decimal separator 3.45 = 3 + 45/100 , comma decimal separator
9 0955 6. 12 = 6 + 12/100 93 = 64 √a square root √a ⋅ √a = a √16 = ±4 909 93 3 √a cube root 3 √a ⋅ 3 √a ⋅ 3 √a = a nth root n √8 = 2 for n =3 % percent 1% = 1/100 10% × 50 = 5 ‰ ppm
Similar Posts |
ML Aggarwal Class 6 Solutions Chapter 10 Basic Geometrical Concept Check Your Progress for ICSE Understanding Mathematics acts as the best resource during your learning and helps you score well in your exams.
## ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 10 Basic Geometrical Concept Check Your Progress
Question 1.
(i) Name all the rays shown in the given figure whose initial point is A.
(ii) Is ray $$\overrightarrow{\mathrm{AB}}$$ different from ray $$\overrightarrow{\mathrm{AD}}$$ ?
(iii) Is ray $$\overrightarrow{\mathrm{CA}}$$ different from ray $$\overrightarrow{\mathrm{CE}}$$ ?
(iv) Is ray $$\overrightarrow{\mathrm{BA}}$$ different from ray $$\overrightarrow{\mathrm{CA}}$$ ?
(v) Is ray $$\overrightarrow{\mathrm{ED}}$$ different from ray $$\overrightarrow{\mathrm{DE}}$$ ?
Solution:
Question 2.
From the given figure, write
(i) all pairs of parallel lines.
(ii) all pairs of intersecting lines.
(iii) lines whose point of intersecting is E.
(iv) collinear points.
Solution:
Question 3.
In the given figure :
(a) Name;
(i) Parallel lines.
(ii) All pairs of intersecting lines.
(iii) concurrent lines.
(b) State wheather true or false:
(i) points A, B and D are collinear.
(ii) lines AB and ED interesect at C.
Solution:
Question 4.
In context of the given figure, state whether the following statements are true (T) or false (F):
(i) Point A is in the interior of ∠AOD.
(ii) Point B is in the interior of ∠AOC.
(iii) Point C is in the exterior of ∠AOB.
(iv) Point D is in the exterior of ∠AOC.
Solution:
Question 5.
How many angles are marked in the given figure? Name them?
Solution:
Question 6.
In context of the given figure, name
(i) all triangles
(ii) all triangles having point E as common vertex.
Solution:
Question 7.
In context of the given figure, answer the following questions:
(i) Is ABCDEFG a polygon?
(ii) How many sides does it have?
(iii) How many vertices does it have?
(iv) Are $$\overline{\mathrm{AB}}$$
and $$\overline{\mathrm{FE}}$$ adjacent sides?
(v) Is $$\overline{\mathrm{GF}}$$
a diagonal of the polygon?
(vi) Are $$\overline{\mathrm{AC}}, \overline{\mathrm{AD}} \text { and } \overline{\mathrm{AE}}$$ diagonals of the polygon?
(vii) Is point P in the interior of the polygon?
(viii)Is point A in the exterior of the polygon?
Solution: |
# Triangle Congruence Calculator
Created by Davide Borchia
Reviewed by Anna Szczepanek, PhD
Last updated: Jan 18, 2024
Congruence is an important concept in geometry: learn how to decide if two triangles are congruent with our triangle congruence calculator.
Here you will learn:
• What congruence is;
• Why congruence in triangles is particular;
• The four main types of congruence in triangles; and
• How to use our triangle congruence calculator.
## What is congruence?
Congruence is a property of geometric shapes (polygons). Take two polygons: we say that they are congruent if all the corresponding sides and angles are congruent.
Why do we need both sides and angles?
• Sides fix the perimeter of a polygon, but not necessarily the angles: think of deforming a square. We say that they fix the scale of the polygon.
• Angles fix the shape but not the scale of a polygon. You don't have information about the size of the shape.
• A combination of both fixes both the shape and the size. We can define congruence only if angles and sizes are defined.
## And what is a triangle?
Triangles are the simplest polygons, with only three sides and three angles. Behind this apparent lack of complexity, they hide many interesting properties and relations, many of them reflected in the way we can analyze their congruence.
We identify the sides of a triangle with the letters $a$, $b$, and $c$, while we mark the angles with the corresponding letters of the Greek alphabet: $\alpha$, $\beta$, and $\gamma$.
## Congruence in triangles: calculate identical triangles.
Triangles are the only shapes that don't require a full set of data to assess their congruence. Some strategies allow you to do so with as little as three quantities. The possible ways to determine if two triangles are congruent are:
• SSS, when three sides are given;
• SAS, when you know two sides and the angle between them;
• ASA, when you know two angles and the side between them; and
• AAS, when you know two angles and one of the adjacent sides.
We will now define the congruence in triangles using these four possible criteria.
#### SSS triangles congruence
You can determine if two triangles are congruent by knowing just the sides. This interesting property stems from the fact that triangles are non-deformable. Apart from making them the most common shape in architecture and engineering, this property allows us to say that if the sides of two triangles are congruent in pairs, then the two triangles are congruent.
J. Hadamard gave the of this theorem — in our opinion.
#### SAS triangles congruence
If two triangles have a pair of congruent sides, and the angle between them has the same amplitude, then they are congruent.
🙋 Euclid proved the SAS triangle congruence theorem using the superposition method, which only involves a "mental" exercise in placing one shape over the other. You can see that this is not a strong mathematical argument, and in fact, the proof is not accepted today. The SAS congruence theorem is often considered a postulate.
#### ASA triangles congruence
An ASA triangle is defined by a set of two angles (say $\alpha$ and $\beta$) and the side between them (in this case, $c$). We can prove the triangles' congruence, but only assuming that the SAS theorem is correct.
Take two triangles: $A \overset{\triangle}{B} C$ and $D \overset{\triangle}{E} F$. We know the angles $A\hat{B}C=D\hat{E}F$ and $C\hat{A}{B}=F\hat{D}E$, and that the side between them is the same in the two triangles: $\bar{AB}=\bar{DE}$.
Assume that the sides $\bar{AC}$ and $\bar{DF}$ are different ($\bar{AC}>\bar{DF}$). We can identify a point $G$ on $\bar{AC}$ where the SAS theorem hold, since $\bar{AG}=\bar{DF}$, $A\hat{B}C=F\hat{D}E$, and $\bar{AB}=\bar{DE}$.
This fixes the other angle ($A\hat{B}C$) to the value of $D\hat{E}F$. At this point you can see that given an ASA set of data, the triangles must be congruent.
#### AAS triangles congruence
If you have two triangles with an identical pair of angles and a congruent side not comprised between the angles, the two triangles are congruent. You just defined an AAS triangle.
You can prove the congruence of a triangle of which you know such data by starting with the ASA triangle proof.
## When size doesn't matter: similar triangles
Pairs of triangles can be compared using angles instead of sides. In this case, we will miss information on the scale of the polygon: an equilateral triangle as big as a galaxy will look identical to an equilateral triangle as small as a molecule if you look from the right distance. This concept is called by mathematicians similarity.
Triangles with the same set of angles are similar. If the sides also coincides, then we deal with congruent triangles.
In an interesting contrast, knowing the sides is enough to calculate the angles, but the inverse is not true: the angles only fix the shape.
## How to use our triangle congruence calculator
Our triangle congruence calculator implements all four possible triangle congruence theorems and crosses them. This means that not only you will be able to compare the same type of input (e.g. SSS and SSS), but also different sets of data (e.g. ASA with SAS): we will perform the needed calculations in the background.
In our calculator, you can see two sets of variables, one for each triangle.
Choose the set of data you know in the variable given for each of them. Proceed to insert the data. If the combination is "legal", we will tell you if the pair of triangles you are studying is congruent or not. If not, we will give you another information: if the triangles you are analyzing are similar.
## Other triangle calculators
This calculator is just one of the many tools about triangles at Omni Calculator. Explore them, starting from this list. You can see many terms you already met in this article!
## FAQ
### Are AAA triangles congruent?
AAA triangles are defined by the three interior angles. AAA triangles are not necessarily congruent. Triangles with the same set of angles have the same shape, but not necessarily the same size: the scale of a shape is not determined by the angles. To fix the shape univocally, you need to set at least one side.
### Is a triangle with sides a = 1, b = 2, and c = 3 congruent with an ASA triangle with ß = 46.57°, γ = 104.48°, and a = 2?
Yes! To check this congruence, you need to calculate some quantities first. You can decide to:
1. Calculate the sides of the ASA triangle; or
2. Calculate the angles of the SSS triangle.
Let's calculate the sides.
1. Compute the angle α:
α = 180° - ß - γ= 28.96°
2. Use the law of sines to compute b and c:
b = (a/sin(α)) × sin(ß) = 3
c = (a/sin(α)) × sin(γ) = 4
3. The two triangles are SSS congruent!
### Is SAS enough to calculate the triangle congruence?
Yes. If you know a set of two sides and the angle between them (SAS), then you can determine if a pair of triangles are congruent. This property has been known since the time of Euclid (he gave a rather unsatisfying proof). To this day, mathematicians accept this theorem as an axiom of geometry.
### Is SAA the same as AAS?
Yes. The AAS postulate to determine the triangle convergence is the same as SAA: the only requirement is to know two angles and one of the adjacent sides. If you knew the last side, you would have to use the ASA postulate.
### Is SSA enough to calculate two triangles congruence?
No. Fixing two sides and one angle gives space to a degree of uncertainty: it is possible to rotate the first side in two positions, one at an acute angle and the other at an obtuse angle. In this case, the scale of the polygon is fixed, but not the shape.
Davide Borchia
Triangle 1
Given
three sides (SSS)
a
cm
b
cm
c
cm
Triangle 2
Given
three sides (SSS)
a
cm
b
cm
c
cm
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# Lesson (I)
## Set theory as a foundation for mathematics
Set theory is a foundation for mathematics. This means
• All abstract mathematical concepts can be expressed in the language of set theory.
• All concrete mathematical objects can be encoded as sets.
• We can write down our mathematical reasoning in set theory: true statements about mathematical objects can be deduced from the axioms of set theory and the rules of logic.
Georg Cantor, the founder of set theory, gave an informal definition of sets:
By a set we mean any collection $M$ of determinate, distinct objects (called the elements of $M$) of our intuition or thought into a whole.
Cantor defined sets to make the idea of infinity, which were objects of fascination for millennia, precise, but soon set theory took a much bigger role in mathematics: It became a foundation for the whole of mathematics. This was due to work of many brilliant set theorists and mathematicians, most notably Bourbaki. All mathematical objects you may see in any undergraduate mathematics course is a set: This includes geometric objects such as manifolds, vector bundles, etc as well as algebraic objects such as monoids, groups, rings, etc.
For us, a set is a collection of elements from a specified universe of discourse. The collection of everything in the universe of discourse is called the universal set, denoted by $\mathcal{U}$.
Given a set $A$ of objects in some universe and an object $a$, we write the proposition $a \in A$ to say that the element $a$ belongs to the set $A$. We write $a \not\in A$ as shorthand for the proposition $\neg (a \in A)$.
## Building Sets
### How to Form Sets
But it is not enough to just have a vague of idea of what sets are, we actually need to know how to form sets. Cantor’s informal characterization suggests that whenever we have some property (aka predicate), $P(x)$, of a domain $X$, we can form the set of elements that have the property $P(x)$. We denote this set by $\{ x \in X \mid P(x) \} \, .$ We read this as “the set of element $x$ of $X$ such that $P(x)$” or as “the set of element $x$ of $X$ which satisfy $P(x)$”.
The notation above is called the set-builder notation. We call the set $$\set{x \in X \mid P(x)}$$ the extension of property/predicate $P$.
In above the predicate $P$ had only one variable (aka parameters), but in general it can have many variables. For instance consider the exmaple $P(x,y) = (x+y=0)$ over real numbers. The set $\{ (x,y) \in \RR \times \RR \mid P(x,y) \}$ is the set of all points in the 2-dim plane which lie on the line described by the equation $y = -x$.
Or we can have a predicate
$C(t,h) = (t < 25 ) \land (t > 15) \land (h > 40 ) \land (h < 60)$
where $t$ is a temperature variable measured in centigrade and $h$ is a humidity variable measured in percentage. Now, we can form the set of all places in the world where the predicate $C(t,h)$ holds:
$\set{ x \in W \mid C(t(x),h(x)) }$
Note that we substituted terms $t(x)$ and $h(x)$ for $t$ and $h$ in $C(t,h)$ (Recall how the substitution of terms in predicates works from the lesson on predicates).
When forming sets using predicates, we can think of the predicate $P(x)$ as a sort of filter or checking mechanism: it selects $x$’s according to the filtering criteria.
Here’s another example: Consider the set $\NN$ of natural numbers and the predicate $E$ over $\NN$, that is
$E(n) = n \text{ is even} .$
We form the set of even numbers as follows:
$\EE = \set{n \in \NN \mid E(n) }$
Sometimes in informal mathematical writing, you see the notation $\set{2,4,6,\ldots}$ for the same set $\EE$. This notation rely on the reader’s ability to infer the pattern being followed and correctly guess the intention of the writer, so use it with caution, or better yet, do not use it!
Here are more examples of sets written in the set builder notation:
• $\set{ n \in \ZZ \mid n \text{ is odd} }$
• $\set{n \in \NN \mid n \text{ is prime} }$
• $\set{n \in \ZZ \mid n \text{ is prime and greater than } 2 }$
• $\set{ n \in \NN \mid n \text{ can be written as a sum of its proper divisors} }$
• $\set{ a \in \RR \mid a \text{ is equal to } 1, 2, 3, \text{ or } \pi }$
• $\set{a \in \RR \mid (a < 0) \lor (a > 0)}$
But the following expressions do not form a set:
• $\set{n \in \ZZ \mid \frac{n}{2} }$
• $\set{z \in \CC \mid (z < 0) \lor (z > 0)}$
#### ✏️ Challenge
Suppose $P$ is a predicate on natural numbers. In below, we are defining sets $A$, $B$, $C$ using the set builder notation. Determine which of the following expressions correctly determines a set, and which ones are incorrect/illegal?
• $A = \set{n \in A \mid \neg P(n)}$
• $B = \set{n \in \ZZ \mid n \in \NN}$
• $C = \set{x \in \RR \mid \sqrt{x} < 2}$
• $D = \set{n \in \NN \mid n \text{ is sum of four squares} \lor \exists n P(n)}$
### Different Notations
Sometimes a set is formed without the domain of the predicate being (explicitly) specified. In such a case, we can think of the domain being the universal set $\mathcal{U}$. Therefore the set
$\set{n \mid n \text{ is a natural number}}$
is a shorthand for the set
$\set{n \in \U \mid n \text{ is a natural number}}$
This sets pick all the elements in the universe which are natural numbers.
An alternate form of set-builder notation uses an expression involving one or more variables to the left of the vertical bar, and the range of the variable(s) to the right. The elements of the set are then the values of the expression as the variable(s) vary: $\set{ \mathsf{expr}(x) \mid x \in X } \text{ is defined to mean } \set{ y \mid \exists x \in X (y = \mathsf{expr}(x) )}$
For example, the expression $\set{2n \mid n \in \NN }$ denotes the set $\EE$ of even numbers. It is shorthand for $\set{n \in \mathbb{N} \mid \exists k \in \mathbb{N},\, n=2k }$.
We can even use a mix of the two notations: The set $\set{ p^2 + 1 \mid p \text{ is prime} }$ consists of all numbers whose predecessor is a square of a prime number.
## Some Important Sets
Using set-builder notation, we can define a number of common and important sets.
• The empty set: $\emptyset = \set{ x \mid \bot } \, .$
• The universe: $\U = \set{ x \mid \top } \, .$
• Singleton sets: for an object $a$, the singleton set $\set{a}$ is formed as $\set{ x \mid x = a }$.
• For distinct objects $a$ and $b$, $\set{ x \in \mathcal U \mid (x = a) \lor (x = b) }$ gives the set $\set{a , b}$.
• Similarly, $\set{ x \in \mathcal U \mid (x = a_1) \lor (x = a_2) \lor \ldots \lor (x=a_n) }$ describes the set contaning $n$ elements $a_1, \ldots, a_n$.
We say that a set $X$ is inhabited if it has at least one element. Formally, a set $X$ is inhabited if the sentence $\exists x \in X. \top$ or equivalently the sentence $\exists x \, (x \in X) )$ is true.
We say that a set $X$ is empty if it is not inhabited, that is the sentence $\neg \exists x \, (x\in X)$ is true. A set is non-empty if it is not empty.
#### ✏️ Challenge
Give a constructive proof that the set $\emptyset$ is empty.
#### ✏️ Challenge
Give a natural deduction proof of the fact that that every inhabited set is non-empty.
We say that a set $A$ is decidable if the sentence $\forall x. x \in A \lor x \notin A$ holds.
## Operations on Sets
Using the set-builder notation, we can define a number of primary operations on sets.
• Union of sets $A$ and $B$ is given by the set $A \cup B = \set{ x \mid x \in A \lor x \in B }$
• Intersection of sets $A$ and $B$ is given by the set $A \cap B = \set{ x \mid x \in A \land x \in B }$
• Complement of a set $A$ is given by the set $\compl{A} = \set{ x \mid \neg (x \in A) }$
• Relative complement of set $A$ with respect to set $B$ is given by the set $A \setminus B = \set{ x \mid x \in A \land \neg x \in B }$
One can define many more operations on sets, but they can be derived from these primary operations. This is similar to the situation with logical connectives. An important example is the disjoint union of two sets, say $A$ and $B$, defined as follows: $A \sqcup B \defeq (A \times \set{0}) \cup (B \times \set{1})$
Sometimes we denote the disjoint union with a different notation, namely “$+$”. For instance, $\set{🥕} + \set{⛄} = \set{🥕,⛄} \quad \text{ and } \quad \set{🥕} + \set{🥕} = \set{ (🥕, 0), (🥕, 1)}$
#### Challenge
What is $\NN + \NN$?
#### Challenge
Describe the elements of $(A + B) + C$ and $A + (B + C)$.
By their definition, the primary operations on sets are intimately related to the logical connectives. In fact one can prove that
• $$\forall x \; (x \in \emptyset \leftrightarrow \bot)$$
• $$\forall x \; (x \in \mathcal U \leftrightarrow \top)$$
• $$\forall x \; (x \in A \cup B \leftrightarrow x \in A \vee x \in B)$$
• $$\forall x \; (x \in A \cap B \leftrightarrow x \in A \wedge x \in B)$$
• $$\forall x \; (x \in \compl A \leftrightarrow \neg x \in A)$$
• $$\forall x \; (x \in A \setminus B \leftrightarrow x \in A \wedge \neg x\in B)$$
The connection between logic and set theory runs even deeper; we make this more clear once we are equipped with the notion of function between sets.
## Equality of Sets
A mathematical principle is that whenever we introduce a mathematical concept, we should ask ourselves what it means for two instances/realizations of that concept to be identified (aka equal). For instance, for the concept of natural number, this is intuitively clear to us: we know without a moment of reflection that $2=2$ but $\neg(2=3)$. However when the concepts we introduce get more complicated judging whether there is an identification (aka an equality) between instances of those concepts becomes less obvious.
Here, we ask ourselves when two sets are equal?
1. Are the sets $\set{n \in \NN \mid \exists k \in \NN,\, n = 2k } \quad \text{and} \quad \set{n \in \QQ \mid \exists k \in \NN,\, n = 2k }$ equal?
2. How about “the set of prime numbers less than $2$” and “the set of even prime numbers greater than $2$”?
3. How about $\set{x \in \QQ \mid x^2 < 2} \quad \text{and} \quad \set{x \in \QQ \mid x^2 \leq 2} \, ?$
We define two sets $A$ and $B$ to be equal precisely when they have the same elements, in other words, every element of $A$ is an element of $B$ and vice versa. We write $A=B$ to denote the equality of $A$ and $B$.
The formal sentence expressing $A=B$ is
$\forall x \, (x \in A \Leftrightarrow x \in B) \, .$ Let us define predicate $P_A$ and $P_B$ by $P_A (x) = (x \in A) \quad \text{and} \quad P_B (x) = (x \in B) \, .$ Therefore, the definition of equality of sets says that $A$ and $B$ are equal sets if and only if $P_A$ and $P_B$ are equivalent predicates. Conversely, starting from equivalent predicates $P$ and $Q$, ranging over a domain $X$, we obtain the equality $\set{x \in X \mid P(x)} = \set{x \in X \mid Q(x)} \, .$
We say that this notion of equality is extensional since although two predicates $P(x)$ and $Q(x)$ can have different intensions they can have the same extension and therefore giving rise to the same sets. Consider for instance the example above where $P(n)$ is defined to be “$n$ is a prime numbers less than $2$” and $Q(n)$ is defined to be “$n$ is an even prime number greater than $2$”.
Therefore, using the extensional definition of equality of sets, the answers to the questions (1)-(3) of the previous slide are all positive.
As an example we prove the distributivity of intersection ($\cap$) over union ($\cup$) of sets.
Theorem. Let $A$, $B$, and $C$ denote sets of elements of some domain. Then $A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \, .$
Proof. Let $x$ be an arbitrary element of $A \cap (B \cup C)$. Then $x$ is in $A$, and either $x$ is in $B$ or $x$ is in $C$. In the first case, $x$ is in $A$ and $x$ is in $B$, and hence $x$ is in $A \cap B$. In the second case, $x$ is in $A$ and $C$, and hence $x$ is in $A \cap C$. Therefore, $x$ is in $(A \cap B) \cup (A \cap C)$.
Conversely, suppose $x$ is in $(A \cap B) \cup (A \cap C)$. There are now two cases:
• First, suppose $x$ is in $A \cap B$. Then $x$ is in both $A$ and $B$. Since $x$ is in $B$, it is also in $B \cup C$, and so $x$ is in $A \cap (B \cup C)$.
• The second case is similar: suppose $x$ is in $A \cap C$. Then $x$ is in both $A$ and $C$, and so also in $B \cup C$. Hence, in this case also, $x$ is in $A \cap (B \cup C)$, as required. $\qed$.
You should be able to translate the proof above into a natural deduction proof, although it may look a bit unruly.
#### Challenge
Construct a natural deduction proof of the sentence $\forall x \; (x \in A \cap (B \cup C) \leftrightarrow x \in (A \cap B) \cup (A \cap C)) \, .$
#### Challenge
1. Prove the following equalities of sets, possibly using the Law of Excluded Middle. Identify the statements whose proof require invoking LEM. Explain why.
$A \cup \compl A = \mathcal U$$A \cap \compl A = \emptyset$
$A \cup A = A$$A \cap A = A$
$A \cup \emptyset = A$$A \cap \emptyset = \emptyset$
$A \cup \mathcal U = \mathcal U$$A \cap \mathcal U = A$
$A \cup B = B \cup A$$A \cap B = B \cap A$
$(A \cup B) \cup C = A \cup (B \cup C)$$(A \cap B) \cap C = A \cap (B \cap C)$
$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$
1. How do you describe the resemblance between the columns?
For sets $A$ and $B$, the set $A$ is said to be a subset of $B$, written $A \subseteq B \, ,$ if every element of $A$ is an element of $B$. In contrast to equality of sets we can think of $\subseteq$ as “inequality” of sets. It is similar to the inequality $2 \leq 3$ (We make this idea more precise later when we will talk about partial orders.)
Formally, $A \subseteq B$ is expressed by the sentence [ \forall x \; (x \in A \To x \in B) \, . ]
#### Challenge
Do you think that $\varnothing \subseteq \varnothing$? Justify your answer by providing a proof or a counter-example.
Theorem. For any two sets $A$ and $B$ we have $A=B$ if and only if $A \subseteq B$ and $B \subseteq A$.
#### Challenge
Prove the following facts about the subset relationship:
• For all sets $A$ we have $A \subseteq A$.
• For all sets $A$, $B$ and $C$, if $A \subseteq B$ and $B \subseteq C$ then $A \subseteq C$.
• For all sets $A$ we have $\emptyset \subseteq A$.
• For all sets $A$, $B$, if $A \cup B = B$ then $A \subseteq B$.
• For all sets $A$, $B$, if $A \cap B = A$ then $A \subseteq B$.
#### Challenge
Which of the following statements are true regardless of the choice of sets $A,B,C$? Justify your answer.
• $A \subseteq B \cup C \implies A \subseteq B \lor A \subseteq C$
• $A \subseteq B \cap C \implies A \subseteq B \land A \subseteq C$
• $A \cap B \subseteq C \implies A \subseteq C \land A \subseteq C$
• $A \cap B \subseteq C \implies A \subseteq C \lor A \subseteq C$
• $A \cup B \subseteq C \implies A \subseteq C \lor A \subseteq C$
#### Challenge
Constructively prove the following inequlities of sets.
$\compl{(A \cup B)} \subseteq \compl{A} \cap \compl{B}$$\compl{(A \cap B)} \supseteq \compl{A} \cup \compl{B}$
$\compl{(A \cup B)} \subseteq \compl{A} \cap \compl{B}$$\compl{(A \cap B)} \supseteq \compl{A} \cup \compl{B}$
## Classical Sets
We can show constructively that for any set $A$, $A \subseteq \compl{\compl{A}}$
#### ✏️ Challenge
Show that the Law of Excluded Middle is equivalent to the statement that for every set $A$, $\compl {\compl{A}} \subseteq A \, .$
### De Morgan for Sets
Since sets are formed as extension of predicates we expect to have the following equalities of sets if we reason classically.
$\compl{(A \cup B)} = \compl{A} \cap \compl{B}$$\compl{(A \cap B)} = \compl{A} \cup \compl{B}$
$\compl{(A \cup B)} =\compl{A} \cap \compl{B}$$\compl{(A \cap B)} = \compl{A} \cup \compl{B}$ |
# Geometry
## Objective
Use equations to solve for unknown angles. (Part 2)
## Common Core Standards
### Core Standards
?
• 7.G.B.5 — Use facts about supplementary, complementary, vertical, and adjacent angles in a multi-step problem to write and solve simple equations for an unknown angle in a figure.
?
• 4.MD.C.5
• 4.MD.C.7
## Criteria for Success
?
1. Identify angle relationships in angle diagrams involving vertical, supplementary, and complementary angles.
2. Write equations to represent relationships between known and unknown angle measurements.
3. Determine the measures of unknown angles and judge the reasonableness of the measures.
## Tips for Teachers
?
• Lessons 3 and 4 present students with more complicated line and angle diagrams where they use various relationships between the angles to set up equations and solve for unknown angle measurements. In Lesson 4, students encounter angles that are described by variables with a coefficient other than 1. They will distinguish between solving for the value of the variable and determining what the measure of the angle is.
• As with Lesson 3, some students may struggle with identifying the relevant relationships and setting up a corresponding equation; others may struggle with solving the equations. Keep an eye out for both/either struggles in order to address incorrect answers appropriately.
• Continue to have students describe what they see in the angle diagrams before they solve. This can support them in understanding the structure of the diagram (identifying key lines or features such as right angles) and in making sense of the diagram by identifying relationships between lines and angles (MP.7 and MP.8).
#### Remote Learning Guidance
If you need to adapt or shorten this lesson for remote learning, we suggest prioritizing Anchor Problem 2 (benefits from worked example). Find more guidance on adapting our math curriculum for remote learning here.
#### Fishtank Plus
• Problem Set
• Student Handout Editor
• Vocabulary Package
## Anchor Problems
?
### Problem 1
For each diagram below, write and solve an equation to determine the value of the variable. Then determine the measure of each angle in the diagrams.
1.
1.
1.
### Problem 2
Two lines meet at a point that is also the endpoint of two rays.
1. Describe the angle relationships you see in the diagram.
2. Set up and solve an equation to find the value of ${x.}$
3. Find the measurements of ${\angle BAC}$ and ${\angle BAH}$.
#### References
EngageNY Mathematics Grade 7 Mathematics > Module 6 > Topic A > Lesson 3Example 3
Grade 7 Mathematics > Module 6 > Topic A > Lesson 3 of the New York State Common Core Mathematics Curriculum from EngageNY and Great Minds. © 2015 Great Minds. Licensed by EngageNY of the New York State Education Department under the CC BY-NC-SA 3.0 US license. Accessed Dec. 2, 2016, 5:15 p.m..
## Problem Set
?
The following resources include problems and activities aligned to the objective of the lesson that can be used to create your own problem set.
Line AC intersects line EB at point F. Ray FD extends from point F. Determine the measures of $\angle EFD$ and $\angle CFD$. |
## What does relative rate of change mean
The calculator will find the average rate of change of the given function on the given interval, with steps shown.
At t equals zero or d of zero is one and d of one is two, so our distance has increased by one meter, so we've gone one meter in one second or we could say that our average rate of change over that first second from t equals zero, t equals one is one meter per second, but let's think about what it is, A concept that will likely be on the midterm but will not be on webwork. An average rate of change tells you the average rate at which something was changing over a longer time period. While you were on your way to the grocery store, your speed was constantly changing. The mathematical definition of slope is very similar to our everyday one. In math, slope is the ratio of the vertical and horizontal changes between two points on a surface or a line. The vertical change between two points is called the rise, and the horizontal change is called the run.The slope equals the rise divided by the run: .
## This function is f (t) = 25 (t – 1) 1/2 and the relative rate of change RRC = f' (t)/f (t). b. Evaluate the What is the meaning and importance of price elasticity.
In any quantitative science, the terms relative change and relative difference are used to When there is no reference value, the sign of Δ has little meaning in the comparison of the two values since it The absolute change in this situation is 1 percentage point (4% − 3%), but the relative change in the interest rate is:. Relative Rate of Change. The relative rate of change of a function f(x) is the ratio if its derivative to itself, namely. R(f(x))=(f^'(x)). SEE ALSO: Derivative, Function, Relative rate of change in given by$\frac{f'(x)}{f(x)}$ If f(x) =$x^2 What does it mean when two currencies are compared in their exchange 17 Nov 2017 It means that the rate of change, relative to the actual value, is constant. It always increases (or decreases) by a fixed percentage in a given time ### Relative change also refers to the change in the indicator in percentage terms, i.e. absolute change as a percentage of the value of the indicator in period 1. Example 1 9,800 workers were made redundant (i.e. retrenched or prematurely released from their contracts) in 2010, compared with 23,430 in 2009. In any quantitative science, the terms relative change and relative difference are used to When there is no reference value, the sign of Δ has little meaning in the comparison of the two values since it The absolute change in this situation is 1 percentage point (4% − 3%), but the relative change in the interest rate is:. Relative Rate of Change. The relative rate of change of a function f(x) is the ratio if its derivative to itself, namely. R(f(x))=(f^'(x)). SEE ALSO: Derivative, Function, Relative rate of change in given by[math] \frac{f'(x)}{f(x)}$ If f(x) =$x^2 What does it mean when two currencies are compared in their exchange 17 Nov 2017 It means that the rate of change, relative to the actual value, is constant. It always increases (or decreases) by a fixed percentage in a given time 28 Jun 2018 f'(t)=17(9e17t) and the relative change is. f'(t)f(t)=17(9e17t)9e17t=17. 6 Aug 2019 You have set δLL= 50 percent which is quite large, as the equation holds in the limit as δL→0. That is, calculating the relative change in volume ### The annual percentage growth rate is simply the percent growth divided by N, the number Another common method of calculating rates of change is the Average To calculate your future balance in the above example the formula would be:. The annual percentage growth rate is simply the percent growth divided by N, the number Another common method of calculating rates of change is the Average To calculate your future balance in the above example the formula would be:. This function is f (t) = 25 (t – 1) 1/2 and the relative rate of change RRC = f' (t)/f (t). b. Evaluate the What is the meaning and importance of price elasticity. A percentage change is a way to express a change in a variable. It represents the relative change between the old value and the new one. For example, if a house is worth 100,000 today and the year after its value goes up to 110,000, the percentage change of its value can be expressed as Relative change also refers to the change in the indicator in percentage terms, i.e. absolute change as a percentage of the value of the indicator in period 1. Example 1 9,800 workers were made redundant (i.e. retrenched or prematurely released from their contracts) in 2010, compared with 23,430 in 2009. Rate of change is used to mathematically describe the percentage change in value over a defined period of time, and it represents the momentum of a variable. The calculation for ROC is simple in that it takes the current value of a stock or index and divides it by the value from an earlier period. ## this section, we will define marginal revenue as the rate of change of the revenue function, even when the revenue A. (For points and secant lines to the left of point A, there would be a similar figure and If x denotes the relative humidity (in . Relative Rate of Change. The relative rate of change of a function f(x) is the ratio if its derivative to itself, namely. R(f(x))=(f^'(x)). SEE ALSO: Derivative, Function, Relative rate of change in given by[math] \frac{f'(x)}{f(x)}$ If f(x) =[math] x^2 What does it mean when two currencies are compared in their exchange 17 Nov 2017 It means that the rate of change, relative to the actual value, is constant. It always increases (or decreases) by a fixed percentage in a given time 28 Jun 2018 f'(t)=17(9e17t) and the relative change is. f'(t)f(t)=17(9e17t)9e17t=17.
Chemical kinetics: the area of chemistry dealing with the speeds or rates at which reactions The speed of a reaction (reaction rate) is expressed as the change in concentration of a reactant or means in the general expression: rate In reactions, most likely, one would be interested in knowing how much reactant is left. 20 Feb 2020 But how can you define the rate at which a reaction occurs? How Do You Determine the Rate of Reaction? Since concentration is measured in molarity, the change in concentration is measured in M while the time is By contrast, changes in reaction rate resulting from hindered diffusion at high conversions the linear volatilization rate constants are sufficient to define the steady-state And, relative rates of reaction are compared by examining rate constants. Although a catalyst increases the rate of a reaction, it does not change the Relative Rate of Change (RRC). [256]. ′ ( ). ( ). (find the derivative of f(x) and divide by f(x)). Also can be found with the dx( ln (f(p)). At what rate is the distance between the cars changing at the instant the second car has been traveling for 1 hour? z x y. Set up the problem by extracting |
# Prime Numbers
Here we will learn about prime numbers, including what they are, how we can determine whether a number is prime, and how to solve problems that involve prime numbers.
There are also prime number worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.
## What are prime numbers?
Prime numbers are positive integers (whole numbers) that have only two factors, themselves and 1.
This means that you cannot divide a prime number by any number apart from 1 or itself, and get an integer answer.
A number that is not prime is called a composite number.
There are 25 prime numbers between 1 and 100. These are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97.
Notice that,
1 is not a prime number as it has only 1 factor.
2 is the only even prime number.
You should be able to recall the first 8 prime numbers (up to 20 ) and be able to determine whether a number is prime by looking at the factors of the number.
To determine whether the number is prime, check whether it has any factors other than itself and 1, either manually or by using a number trick. If the number has a factor that is not itself or 1, it is not prime.
There are a few useful number tricks that can help us determine whether a number is divisible by an integer and therefore has that integer as a factor:
Once we have considered these number tricks, we need to look for divisibility by other numbers. We don’t, however, need to consider every number.
The reason for this is that for a number to be divisible by 4 or 6 or any multiple of 2, it must also be divisible by 2, which is something we have already checked.
For a number to be divisible by 6 or 9 or any multiple of 3, it must also be divisible by 3, which is something that we have already checked.
This pattern continues and means that we only need to check for divisibility by other prime numbers, starting with the number 7.
## How to find prime numbers
In order to determine whether a number is a prime number:
1. Use the number tricks to see whether \bf{2, 3} or \bf{5} is a factor.
2. If they are not factors, test for divisibility by other prime numbers, starting with \bf{7}.
3. State your conclusion with a reason.
### Related lessons onfactors, multiples and primes
Prime numbers is part of our series of lessons to support revision on factors, multiples and primes. You may find it helpful to start with the main factors, multiples and primes lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:
## Prime numbers examples
### Example 1: two digit number
Is 53 a prime number?
1. Use the number tricks to see whether \bf{2, 3} or \bf{5} is a factor.
The last digit is not a 2, 4, 6, 8, or 0 so it is not a multiple of 2.
Adding the digits together we have 5+3=8 and so it is not a multiple of 3.
The last digit is not a 5 or a 0 and so it is not a multiple of 5.
2If they are not factors, test for divisibility by other prime numbers, starting with \bf{7}.
We need to continue dividing 53 by successive prime numbers until we reach the first prime number that is greater than \sqrt{53}.
\sqrt{53} \approx 7.3 so we only need to try prime numbers less than this.
Start by dividing by 7:
As 53\div{7}=7.57142857 \ 7 is not a factor of 53.
The next prime number is 11, which is greater than 7.3 so there are no more integers that we need to try.
3State your conclusion with a reason.
53 is a prime number as it only has two factors, 1 and 53.
### Example 2: three digit number
Is 223 a prime number?
Use the number tricks to see whether \bf{2, 3} or \bf{5} is a factor.
If they are not factors, test for divisibility by other prime numbers, starting with \bf{7}.
State your conclusion with a reason.
### Example 3: four digit number
Is 2073 a prime number?
Use the number tricks to see whether \bf{2, 3} or \bf{5} is a factor.
If they are not factors, test for divisibility by other prime numbers, starting with \bf{7}.
State your conclusion with a reason.
### Example 4: determine the prime number from a list of numbers
One of the following numbers is prime. Identify the prime number.
Use the number tricks to see whether \bf{2, 3} or \bf{5} is a factor.
If they are not factors, test for divisibility by other prime numbers, starting with \bf{7}.
State your conclusion with a reason.
## Problem solving with prime numbers
In order to place numbers into a diagram / table:
1. Determine what each box / section of the diagram represents.
2. Place each number one-by-one into the diagram.
3. Determine the frequency of values in each box / section if required.
## Problem solving with prime numbers examples
### Example 5: complete the diagram (two way table)
Place the set of numbers N=\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} into the two-way table below.
Determine what each box / section of the diagram represents.
Place each number one-by-one into the diagram.
Determine the frequency of values in each box / section if required.
### Example 6: complete the diagram (Venn diagram)
The Venn diagram below represents the two sets P=\{ Prime \} and F=\{ Factor of 24 \}.
Given that ξ=\{x is an integer such that 0 < x \leq 24 \}, complete the Venn diagram.
Use your Venn diagram to write down the frequency of values in the sets.
i) P
ii) P \cap F
Determine what each box / section of the diagram represents.
Place each number one-by-one into the diagram.
Determine the frequency of values in each box / section if required.
### Common misconceptions
• \bf{1} is not a prime number
1 is not a prime number. This is because it only has one factor, rather than the 2 factors needed to be a prime number.
• \bf{2} is a prime number
2 is the only even prime number. This is because it only has two factors, 1 and itself, and therefore by definition, it is a prime number. Every other even number is divisible by 2. \ 2 is therefore known as a special case when discussing prime numbers.
• Prime numbers and the odd numbers
As all but one of the prime numbers are odd (remember that 2 is the only even prime number), it is sometimes assumed that all odd numbers are prime. Take the number 9. It is an odd number, however 9 is a multiple of 3 and so we can divide 9 by 3 (and get 3 ). Not all odd numbers are prime, and not all prime numbers are odd.
• Decimals cannot be prime numbers
All prime numbers are positive integers but a common misconception is that decimals can be prime. For example, the decimal 2.3 is considered to be a prime number as 23 is a prime number.
### Practice prime numbers questions
1. Determine what type of number 71 is from the list below.
Multiple of 3
Multiple of 4
Multiple of 5
Prime number
71 does not end in 0, 2, 4, 6 or 8 and so is not a multiple of 2. It cannot therefore be a multiple of 4.
7+1=4 so 71 is not a multiple of 3 .
71 does not end in 0 or 5 therefore it is not a multiple of 5.
71 \div 7=10.14 therefore it is not a multiple of 7.
This tells us that 71 is a prime number.
2. Determine what type of number 513 is from the list below.
Multiple of 2
Multiple of 3
Multiple of 5
Prime number
513 is a multiple of 3 as 5+1+3=9 which is a multiple of 3.
513\div{3}=171
3. Determine what type of number 143 is from the list below.
Multiple of 3
Multiple of 7
Multiple of 11
Prime
1+4+3=8 so 143 is not a multiple of 3 .
143\div 7=20.42857… so 143 is not a multiple of 7 .
143\div{11}=13 so 143 is a multiple of 11 .
4. Which number from the list is prime?
81
343
2
3\times{10}^{2}
2 is the only even prime number. It has two factors, 1 and itself.
81 has factors of 1, 3, 9, 27 and 81.
343 has factors of 1, 7, 49 and 343.
3\times{10}^{2} written as an ordinary number is 300 which has many factors.
5. Let N=\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}. Which two way table correctly sorts the set N?
Factors of 12: 1, 2, 3, 4, 6, 12
Prime Numbers: 2, 3, 5, 7, 11
6. The Venn diagram below represents the two sets P=\{ Prime \} and F=\{ Factor of 10\}. Given that ξ= \{x is an integer such that 0 < x \leq 12 \}, work out the frequency of values that would appear in P \cap F.
1
2
2 and 5
10
Factors of 10: 1, 2, 5, and 10.
Prime Numbers: 2, 3, 5, and 7.
Each value placed correctly in the Venn diagram gives the result:
The frequency that corresponds to each set is therefore:
### Prime numbers GCSE questions
1. (a) Write the first 8 prime numbers.
(b) Peter says “every prime number is odd”. Is Peter correct? Explain your answer.
(3 marks)
(a)
2, 3, 5, 7, 11, 13, 17, 19
(1)
(b)
No
(1)
2 is the only even prime number.
(1)
2. (a) The perimeter of a rectangle can be written as P=2(l+w) where l and w are integer side lengths.
Given that the area of the rectangle is 23cm^{2}, calculate the perimeter of the rectangle.
(b) A square has an area of A=x^{2}, where x is an integer. Is the area always, sometimes or never a prime number? Explain your answer.
(4 marks)
(a)
l=1 and w=23 or l=23 and w=1
(1)
P=2(23+1)=2\times{24}=48\text{cm}
(1)
(b)
All square numbers have an odd number of factors, whereas all prime numbers have 2 factors.
or
1 is not prime and any other square number has a factor of its square root meaning it is not prime.
(1)
Never
(1)
3. Below is a list of numbers:
(a) Which of the values is a prime number?
(b) Which of the values is 2 more than a prime number, and 2 less than another prime number?
(c) Which of the values gives a prime number when it is squared?
(3 marks)
(a) 73
(1)
(b) 21
(1)
(c) \sqrt{5}
(1)
4. (a) On the Venn diagram below, P=\{ Prime numbers \} and M=\{ Multiples of 4\}.
For the set of numbers ξ=\{x such that 20 < x \leq 40 \}, complete the Venn diagram.
(b) Write down the numbers that are in the set P .
(c) How many numbers are in the set P \cap M?
(6 marks)
(a)
Prime Numbers: 23, 29, 31, 37.
(1)
Multiples of 4: 20, 24, 28, 32, 36, 40.
(1)
No values in the intersection (P \cap M).
(1)
(P \cup M)’: 21, 22, 25, 26, 27, 30, 33, 34, 35, 38, 39.
(1)
(b) 23, 29, 31, 37
(1)
(c) 0
(1)
## Learning checklist
You have now learned how to:
• Identify prime numbers
• Know and use the vocabulary of prime numbers, prime factors and composite (non-prime) numbers
• Establish whether a number up to 100 is prime and recall prime numbers up to 19
## The next lessons are
The sieve of Eratosthenes |
# GRE Math : How to find percentage from a fraction
## Example Questions
### Example Question #1 : How To Find Percentage From A Fraction
If one of the employees across both industries were to be selected at random, what is the probability that the employee will be a construction industry worker who stayed in the same role for 5 years or more?
17%
26%
65%
20%
10%
26%
Explanation:
The first step is to figure out the percentage of construction employees that have stayed in the same role for 5 years or more—this would include both the "5 to 9 year" and "10+ years" ranges. This would be 0.25 + 0.4 = 65% of all construction employees. To convert to the number of employees, we take the percentage of their total, 0.65 * 8,000,000 = 5,200,000 workers. However, since the probability we are attempting to find is of workers between both industries, we must add the 8 million to the 12 million = 20 million workers total. 5,200,000/20,000,000 = 0.26, or a 26% chance.
### Example Question #2 : How To Find Percentage From A Fraction
For every 1000 cookies baked, 34 are oatmeal raisin.
Quantity A: Percent of cookies baked that are oatmeal raisin
Quantity B: 3.4%
The two quantities are equal.
Quantity B is greater
Quantity A is greater
The relationship cannot be determined from the information given.
The two quantities are equal.
Explanation:
Simplify Quantity A by dividing the number of oatmeal raisin cookies by the total number of cookies to find the percentage of oatmeal raisin cookies. Since a percentage is defined as being out of 100, either multiply the resulting decimal by 100 or reduce the fraction until the denominator is 100. You will find that the two quantities are equal.
### Example Question #3 : How To Find Percentage From A Fraction
50 students took an exam. There were 4 A's, 9 B's, 15 C's, 8 D's, and the rest of the students failed. What percent of the students failed?
Explanation:
students failed.
which equals 28%. |
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Problem 18
# The circles $$x^{2}+y^{2}+2 a x-c^{2}=0$$ and $$x^{2}+y^{2}+2 b x-c^{2}=0$$ intersect at $$A$$ and B. A line through $$A$$ meets one circle at $$P$$ and a parallel line through $$B$$ meets the other circle at Q. Show that the locus of the mid point of $$P Q$$ is a circle.
Expert verified
Hence, it has been proven that the locus of the midpoint of $$PQ$$ is a circle.
See the step by step solution
## Step 1: Find Intersection Points
Since circles intersect at $$A$$ and $$B$$, we can find the intersection points by setting the equations of the circles equal to each other. This gives $$2 a x - 2 b x = 0$$ implying $$x = 0$$. Substituting x = 0 in any of the circle equations will give $$y = c$$ and $$y = -c$$. Therefore the intersection points $$A$$ and $$B$$ are $$(0,c)$$ and $$(0,-c)$$ respectively.
## Step 2: Deduce Properties of Parallel Lines Intersecting Circles
By the nature of parallel lines, point $$P$$ on one circle and matched point $$Q$$ on the other not only form a similar structure but are mirror images with respect to the x-axis. Therefore, if $$P$$ has coordinates $$(x_{1}, y_{1})$$, then $$Q$$ will have coordinates $$(x_{1}, -y_{1})$$. Also, $$P$$ and $$Q$$ lie on the given circles, so they satisfy their equations.
## Step 3: Compute coordinates of Midpoint of $$PQ$$
The midpoint between two points $$(x_{1}, y_{1})$$ and $$(x_{2}, y_{2})$$ is given by $$\left( \frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}\right)$$. Using this formula, the midpoint of $$PQ$$ will have coordinates $$\left(\frac{x_{1} + x_{1}}{2}, \frac{y_{1} - y_{1}}{2}\right)$$, simplifying this gives $$\left(x_{1}, 0\right)$$ which is a point on the x-axis.
## Step 4: Show that Locus is a Circle
A locus of points equidistant from a set point is a circle. In this case, the midpoints of $$PQ$$ are all on the x-axis ($$y = 0$$) and spread symmetrically around the origin. Their x-coordinates satisfy the equation of a circle centered at origin with radius $$\frac{a+b}{2}$$, which is $$x^{2} - (\frac{a+b}{2})^{2}=0$$. Hence the locus is a circle.
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# How to find b squared
Spis treści
## The Definition of b Squared
b squared is a mathematical term that refers to the result of multiplying a number, b, by itself. In other words, it is the product obtained when b is multiplied by itself. This operation can also be expressed as b raised to the power of 2 or written as b^2.
Understanding the definition of b squared is fundamental in various areas of mathematics and science. It serves as a building block for many concepts and calculations. For example, in geometry, knowing how to calculate the area of a square requires finding the value of one side squared. Similarly, in physics and engineering, equations often involve squaring variables to represent relationships between different quantities.
To find b squared using algebraic manipulation, you simply multiply b by itself: (b)(b) = b^2. This process allows you to determine the value of any variable or constant raised to the power of 2 quickly and efficiently.
By grasping this concept and its applications across different disciplines, individuals can enhance their problem-solving skills and gain deeper insights into various mathematical principles. Whether it’s calculating areas or understanding complex formulas in science and engineering fields, having a solid understanding of what it means for something to be squared lays an essential foundation for further exploration into these subjects.
## Multiplying b by Itself: A Simple Approach
To calculate b squared, a simple approach is to multiply b by itself. This means taking the value of b and multiplying it by itself. For example, if b is equal to 3, then 3 squared would be calculated as 3 multiplied by 3, which equals 9.
This method can be easily understood and applied in various mathematical problems. Whether you are working with whole numbers or decimals, the process remains the same. By multiplying a number by itself, you obtain its square value.
It’s important to note that when using this approach, it is crucial to correctly identify the value of b before performing the multiplication. Accuracy in determining the correct value will ensure accurate results in calculating b squared.
By following this simple approach of multiplying b by itself, you can quickly determine the square value of any given number and apply it effectively in various mathematical contexts without relying on complex formulas or operations.
## Utilizing the Exponentiation Operation to Find b Squared
To find the value of b squared, we can utilize the exponentiation operation. The exponentiation operation involves raising a number to a certain power, and in this case, we want to raise b to the power of 2. This means that we need to multiply b by itself.
For example, if we have b = 3, then finding b squared would involve multiplying 3 by itself: 3 * 3 = 9. So in this case, b squared is equal to 9.
The process of utilizing the exponentiation operation to find b squared is quite straightforward. We simply take the value of b and multiply it by itself using multiplication or an appropriate mathematical function on a calculator or computer software. By doing so, we obtain the result which represents the square of our initial value for b.
It’s important to note that finding values raised to higher powers follows similar principles but with different exponents. Understanding how exponents work is fundamental in various areas such as algebra and calculus where these operations are frequently used for solving equations or analyzing functions.
• To find b squared, we utilize the exponentiation operation.
• The exponentiation operation involves raising a number to a certain power.
• In this case, we want to raise b to the power of 2.
• This means that we need to multiply b by itself.
• For example, if b = 3, then b squared would be 3 * 3 = 9.
• So in this case, b squared is equal to 9.
• Utilizing the exponentiation operation is straightforward for finding b squared.
• We simply take the value of b and multiply it by itself using multiplication or an appropriate mathematical function on a calculator or computer software.
• By doing so, we obtain the result which represents the square of our initial value for b.
• It’s important to note that finding values raised to higher powers follows similar principles but with different exponents.
• Understanding how exponents work is fundamental in various areas such as algebra and calculus where these operations are frequently used for solving equations or analyzing functions.
## Square Root of b Squared: A Mathematical Principle
The square root of b squared is a fundamental mathematical principle that involves finding the value of b when it is raised to the power of 2 and then taking the square root of that result. It can be represented as √(b^2). This operation allows us to determine the original value of b, which may have been either positive or negative.
Sprawdź to ➡ ➡ How to Find Hamming Distance: A Comprehensive Guide
When we take the square root of b squared, we are essentially reversing the process of squaring a number. By doing so, we obtain both positive and negative roots for any given value of b. This is because squaring a negative number yields a positive result, while squaring a positive number also gives us a positive outcome.
Understanding this mathematical principle is crucial in various areas, such as geometry and trigonometry. In geometric applications, knowing how to calculate the square root of b squared enables us to find lengths or distances within shapes and figures accurately. Additionally, in trigonometry, this concept helps us solve equations involving right triangles and their side lengths.
By grasping the concept behind finding the square root of b squared, individuals can apply it to real-life situations in science and engineering fields. For instance, engineers often need to determine unknown quantities based on known measurements using formulas that involve squares or square roots. Having a solid understanding of this principle allows them to make accurate calculations and design efficient solutions for practical problems without confusion or errors.
## Employing Algebraic Manipulation to Determine b Squared
Algebraic manipulation is a powerful tool in mathematics that allows us to simplify and solve complex equations. When it comes to determining b squared, we can employ various algebraic techniques to find the value of this expression. One approach involves expanding and simplifying expressions involving b.
To determine b squared using algebraic manipulation, we start with the expression (b + c)^2. By applying the distributive property and squaring each term within the parentheses, we obtain b^2 + 2bc + c^2. However, if we are solely interested in finding b squared, we can isolate this term by subtracting 2bc and c^2 from both sides of the equation.
Another method for determining b squared through algebraic manipulation is by utilizing identities such as (a – b)(a + b) = a^2 – b^2. Here, if we let a = b and substitute into the identity, we get (b – b)(b + b) = 0(b + 3b), which simplifies to -4b^2 = 0. Solving for b squared gives us an answer of zero.
By employing these algebraic manipulations, we can easily determine the value of b squared in various scenarios. Whether it’s expanding expressions or applying identities, understanding how to manipulate equations enables us to solve problems efficiently and accurately without relying on guesswork or trial-and-error methods commonly used before the advent of algebraic techniques.
## Utilizing a Calculator or Computer Software for Efficiency
When it comes to calculating b squared, utilizing a calculator or computer software can greatly enhance efficiency. These technological tools are designed to perform complex mathematical operations quickly and accurately. By inputting the value of b into the appropriate function or formula, the result of b squared can be obtained in a matter of seconds.
One advantage of using a calculator or computer software is that it eliminates human error. Manual calculations can often lead to mistakes due to miscalculations or transcription errors. With a calculator or software program, there is less room for these types of errors as they are specifically programmed to follow mathematical rules and formulas precisely.
Additionally, calculators and computer software provide convenience and speed. In situations where multiple values need to be squared, manually performing each calculation can be time-consuming and tedious. However, with the use of technology, these computations can be done almost instantaneously by simply entering the values into the program.
In summary,
utilizing a calculator or computer software for calculating b squared offers numerous benefits such as accuracy, convenience, and speed. These tools eliminate human error while providing quick results for even complex calculations involving multiple values. Whether in academic settings or real-life applications like engineering and science fields, relying on technology ensures efficient computation without compromising accuracy.
## The Importance of Understanding b Squared in Geometry
Geometry is a branch of mathematics that deals with the properties and relationships of shapes, sizes, and spatial configurations. In this field, understanding b squared holds significant importance. By knowing the value of b squared in geometry, we can accurately determine various measurements and make precise calculations.
One crucial application of b squared in geometry is in finding the area of a square or rectangle. The formula for calculating the area involves multiplying the length (b) by its width (also represented as b). By squaring one side of these geometric figures, we obtain a more accurate measurement that allows us to determine their total surface area.
Furthermore, understanding b squared helps us solve problems involving right triangles. The Pythagorean theorem states that in any right triangle, the square of the hypotenuse (c) is equal to the sum of squares on its two legs (a and b). By identifying which side represents 'b’ in a given triangle, we can apply this principle to find missing lengths or verify if a set of sides satisfies this fundamental relationship.
In summary, grasping the concept and significance behind b squared greatly enhances our ability to work with geometric shapes effectively. It enables us to calculate areas accurately and utilize important principles like the Pythagorean theorem when dealing with right triangles. Whether it’s measuring surfaces or solving intricate geometrical problems, comprehending b squared plays an essential role in mastering geometry concepts and applications.
## Real-life Applications of b Squared in Science and Engineering
Real-life Applications of b Squared in Science and Engineering
One practical application of b squared in science is in the field of physics, particularly when calculating the area or volume of objects. For example, when determining the surface area of a cube with side length b, we can use the formula A = 6b^2. This equation allows us to find an accurate measurement for various physical properties, such as heat transfer rates or material requirements.
Sprawdź to ➡ ➡ How to find molar mass with grams
In engineering, b squared plays a significant role in structural analysis and design. Engineers often use this mathematical concept to calculate moments of inertia for different shapes and structures. By knowing the value of b squared, engineers can accurately predict how materials will respond under applied loads and make informed decisions about their designs.
Another real-life application where b squared is essential is in computer graphics and image processing. When working with pixels on a digital screen or manipulating images using software algorithms, calculations involving squares are frequently used. These calculations help determine pixel intensities or adjust color values within an image to achieve desired effects.
These examples highlight just a few instances where understanding and utilizing b squared have practical implications across various scientific and engineering disciplines. By recognizing its significance beyond theoretical mathematics, professionals can apply this concept effectively to solve real-world problems and improve technological advancements without relying solely on computational tools like calculators or computers.
## Exploring the Relationship between b Squared and Other Mathematical Concepts
The concept of b squared is closely related to other mathematical concepts, such as exponents and powers. When we square a number or variable, we are essentially multiplying it by itself. This operation can be represented using the exponentiation notation, where b squared is written as b^2. It is important to understand this relationship in order to solve more complex equations and problems involving exponents.
In geometry, the concept of b squared plays a crucial role in determining the area of squares and rectangles. The area of a square with side length b can be calculated by simply squaring the value of b (b^2). Similarly, for rectangles with sides measuring b and c, their combined area can be found by multiplying these two values (bc), which can also be expressed as (b)(c) or (b+c)^2 – bc.
Furthermore, exploring the relationship between b squared and other mathematical concepts allows us to delve into advanced topics like quadratic equations. Quadratic equations involve terms that are raised to the power of 2, making them directly connected to the concept of squaring numbers or variables. Understanding how these different mathematical ideas intertwine provides a solid foundation for tackling more challenging algebraic problems.
By examining how b squared relates to various mathematical concepts like exponents, geometry, and quadratic equations, we gain a deeper understanding of its significance within mathematics as a whole. This exploration not only enhances our problem-solving skills but also enables us to see connections between seemingly unrelated topics in mathematics.
## Common Mistakes to Avoid When Calculating b Squared
One common mistake when calculating b squared is forgetting to properly square the value of b. It’s important to remember that squaring a number means multiplying it by itself. For example, if b is equal to 3, then b squared would be 3 times 3, which equals 9. Failing to perform this multiplication correctly can lead to inaccurate results.
Another mistake to avoid is confusing the concept of squaring with taking the square root. While finding the square root of a number involves determining what number multiplied by itself equals that given number, squaring a number simply means multiplying it by itself. Mixing up these two operations can result in incorrect calculations and confusion.
Additionally, overlooking negative values of b can also lead to errors when calculating b squared. It’s crucial to consider both positive and negative values for b since squaring any real number will always yield a positive result. Neglecting this aspect may cause miscalculations and hinder accurate mathematical solutions.
By being mindful of these common mistakes – such as not properly squaring the value of b, confusing squaring with taking the square root, and neglecting negative values – you can ensure more accurate calculations when determining b squared in various mathematical contexts without encountering unnecessary errors or complications.
### What is the definition of b squared?
b squared refers to the mathematical operation of multiplying the value of b by itself.
### How can I calculate b squared using a simple approach?
To calculate b squared, you can simply multiply the value of b by itself.
### Is there another method to find b squared?
Yes, you can also utilize the exponentiation operation by raising b to the power of 2 to find b squared.
### What is the square root of b squared?
The square root of b squared is simply the absolute value of b.
### Can algebraic manipulation be used to determine b squared?
Yes, you can employ algebraic manipulation to find b squared by simplifying expressions or solving equations involving b.
### Is it recommended to use a calculator or computer software for calculating b squared?
Yes, using a calculator or computer software can provide efficiency and accuracy in calculating b squared, especially for complex values of b.
### Why is it important to understand b squared in geometry?
Understanding b squared is crucial in geometry as it allows you to calculate the areas of squares and find the lengths of sides in various geometric shapes.
### Are there real-life applications of b squared in science and engineering?
Yes, b squared is frequently used in real-life applications such as calculating areas, volumes, and performing mathematical modeling in science and engineering.
### How is b squared related to other mathematical concepts?
B squared is related to concepts such as multiplication, exponentiation, square roots, and algebraic manipulation, forming the foundation for various mathematical operations.
### What are some common mistakes to avoid when calculating b squared?
Some common mistakes to avoid when calculating b squared include incorrectly squaring negative numbers, forgetting to multiply b by itself, and misinterpreting the concept of b squared in equations or formulas. |
# Converting from one form to another Vertex or Root form to Standard
```Converting from one form to another
Vertex or Root form to Standard - just FOIL and multiply:
Example: y = ½(x – 2)(x + 4) y = ½(x + 2x – 8) y = ½x + x – 4 (Standard form)
Standard to Vertex form – You can (a) complete the square or (b) find the axis of symmetry (x value of the vertex) and
plug x back into the equation to find the y value of the vertex.
(a) Completing the square:
2
Example: y = 3x – 12x – 9
a = 3, b = -12, c = -9 (y-intercept)
2
Step 1: Substitute 0 for y
0 = 3x – 12x – 9
Step 2: Move the constant
9 = 3x – 12x
Step 3: Factor-out the leading coefficient
9 + ___ = 3(x – 4x + ___ )
Step 4: Add (b/2)2 to both sides
Step 5: Factor the perfect square
Step 6: Move the constant
9 + _12_ = 3(x – 4x + _4_ ) Note: we added 3 x 4
21 = 3 (x – 2)2
y = 3 (x – 2)2 – 21
(Vertex form)
2
2
2
(b) Find the axis of symmetry:
2
Example: y = 3x – 12x – 9
a = 3, b = -12, c = -9 (y-intercept)
To find the vertex, use x = -b/2a: Here, x = -(-12)/2(3) = 2, so x = 2 is the Axis of Symmetry and the x-value of our
vertex.
Now, we can replace x with 2 in our equation to find the y-value of our vertex:
2
y = 3(2) – 12(2) – 9 = 12 – 24 – 9 = - 21
Now, we know the value of “a” (3) and our vertex (2,-21), so we can put this in Vertex form:
2
y = 3(x – 2) – 21
Pretty cool, huh?
Vertex to Roots form: When we cannot factor the Standard form to get to the Roots form, we can set y = 0 and use
the Vertex form to solve for our Roots (where the parabola crosses the x-axis):
2
Example:
2
y = 3(x – 2) – 21 0 = 3(x – 2) – 21
2
21 = 3(x – 2)
Divide both sides by 3
2
7 = (x – 2) Take the square root of both sides
±√7 = x – 2 Add 2 to both sides and you have the Roots!
2±√7 = x So y = 3(x – 2+ √7)(x – 2 – √7) (Root form)
Remember: Roots can be imaginary!
2
Example:
y = 2(x + 1) + 2
2
0 = 2(x + 1) + 2
2
-2 = 2(x + 1)
Divide both sides by 2
2
-1 = (x + 1) Take the square root of both sides
Since we cannot take the square root of a negative number, our answer is the imaginary root of -1±i
``` |
# How to solve square roots
It’s important to keep them in mind when trying to figure out How to solve square roots. We can solve math word problems.
## How can we solve square roots
One of the most important skills that students need to learn is How to solve square roots. A triangle solver is a useful tool for finding the area of a triangle. It works by taking into account the size of each side and then comparing them to each other to find the average size of each side. The calculation can be done in one of two ways: either treating the sides as equal, or by calculating the difference between the three measurements. The latter method is more accurate and less prone to rounding error, but it’s also more complex. In most cases, calculating the difference is not necessary and just treating both sides as equal will suffice. However, if you have very small sides that are difficult to measure accurately, you may want to consider using this option. • Solving triangles by area: This method requires determining the area of each triangle’s base. To do this, multiply each side’s length (in centimeters) by its corresponding value from the table below (to convert values into inches, divide by 25.4). Subtract these results from 100. The result is the total base area (in square centimeters). Next, use a calculator to find the area of the triangle’s height (in square centimeters). Finally, use a formula to find the total area of all three triangles (in square centimeters). • Solving triangles by height: This method involves finding the difference between each side’s height (in centimeters),
Geometric sequence solvers are algorithms that can be used to determine the shortest path between two points in a graph. They are widely used in computer science, engineering, and physics. There are two types of geometric sequence solvers: graph traversal methods and graph coloring methods. Graph traversal methods start from the first node and move along all the edges to find the shortest path between any two nodes in the graph. Graph coloring methods start from a given colored vertex and use a specified algorithm to color all the neighboring vertices with different colors. Geometric sequence solvers can be classified into three groups based on how they solve optimization problems: heuristic methods, greedy methods, and branch-and-bound methods. In heuristic methods, an initial hypothesis is tested against each node in the graph to determine whether it is the shortest path between any two nodes. If so, then its length is determined. Otherwise, new hypotheses are generated until a final solution is found. In greedy methods, an initial solution is chosen arbitrarily and then modified if possible to reduce its cost by taking advantage of local optima. In branch-and-bound methods, an initial solution is chosen arbitrarily but then modified according to a heuristic or other criteria until it has been optimized to within an acceptable amount of error. Graph coloring methods are popular because they can be used to find both optimal solutions and approximate solutions for
Quadratic formula is a mathematical formula used to solve for the value of the unknown term in a quadratic equation. The formula can be used to solve for an unknown term in a quadratic equation that involves two known terms. The formula is written as follows: You can also use an online calculator to calculate the value of the unknown term. To solve for the value of the unknown term, do one of the following: If you know the value of the first unknown term, then plug it into one side of the formula and see what results. If you know the value of one of the known terms, then plug that into one side of the formula and see what results. If both sides match, then your answer is correct. If neither side matches, then you need to find another way to solve for that term.
The disparities between minority groups and the majority is a major problem in the United States. Exact statistics on how many minorities are unemployed and how many people of lower income are living in poverty are hard to track, but it’s clear that there is still much to be done. One way that the inequality gap can be closed is by encouraging more minorities to go into STEM fields. This will not only help them to earn more money, but it will also give them more recognition in the workplace and make it easier for them to get raises and promotions. Another way that inequality can be closed is by improving access to education. If more minorities have access to quality education, they will be less likely to end up stuck in low-paying jobs or trapped in poverty.
For students who are new to mathematics, it can be difficult to understand concepts such as variables, formulas and variables. When you're working on a math problem, you might not understand what you're trying to solve or why you should even be solving the problem in the first place. This can be frustrating for both students and teachers. One way to combat this is by using problem-solving tools. These can be visual tools like a worksheet or graph, or they can simply involve posing a question that makes sense from the beginning. For example, when working with a basic addition problem, it might make sense to start by thinking about how much money you have. This will help you determine whether you have enough money to pay for your purchase. You might also think about what things cost in your area, which will help you figure out if it's possible to make the purchase without going into debt.
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Study Materials
# NCERT Solutions for Class 11th Mathematics
## Chapter 4. Principle Of Mathematical Induction
### Exercise 4.1
Chapter 4. Principle of Mathematical induction
Exercise 4.1
Prove the following by using the principle of mathematical induction for all ∈ N:
Solution:
Let the given statement be P(n), i.e.,
LHS = RHS
Thus P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n.
Solution:
Let the given statement be P(n), i.e.,
LHS = RHS
Thus P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n.
Solution: Let the given statement be P(n), so
Thus P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n.
Q19. n (n + 1) (n + 5) is a multiple of 3.
Solution:
Let the given statement be P(n), so
P(n) : n (n + 1) (n + 5) is a multiple of 3.
For n = 1, so we have;
n (n + 1) (n + 5) = 1 × 2 × 6 = 12 = 3 × 4
P(n) is true for n = 1
Assume that P(k) is also true for some positive integer k.
k(k + 1) (k + 5)
= k3 + 6k2 + 5 k = 3m (say) ……………….. (1)
Now, we shall prove that P(k + 1) is true whenever P(k) is true
Replacing k by k + 1
k + 1 (k + 2) (k + 6)
= (k + 1) (k2 + 8k + 12)
= k (k2 + 8k + 12) + 1(k2 + 8k + 12)
= k3 + 8k2 + 12k + k2 + 8k + 12
= k3 + 9k2 + 20k + 12
=( k3 + 6k2 + 5 k) + 3k2 + 15k + 12
= 3m + 3k2 + 15k + 12 from (1)
= 3(m + k2 + 5k + 4)
∴ k + 1 (k + 2) (k + 6) is multiple of 3
Thus P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n ∈ N.
Q20. 102n - 1 + 1 is divisible by 11.
Solution:
Let the given statement be P(n), so
P(n) : 102n - 1 + 1 is divisible by 11.
For n = 1, so we have;
102n - 1 + 1 = 102×1 - 1 + 1 = 10 + 1 = 11
P(n) is true for n = 1
Assume that P(k) is also true for some positive integer k.
102k- 1 + 1 = 11m say
102k- 1 = 11m - 1 ……………… (1)
We shall prove that P(k + 1) is true whenever P(k) is true
replacing k by k + 1 we have
102k - 1 + 1
= 102k + 1 + 1
= 102k × 101 + 1
= {102k - 1 × 100 + 1}
= {(11m - 1)× 100 + 1} from equation (1)
= 1100m - 100+ 1
= 1100m - 99
= 11(100m - 9)
102n - 1 + 1 is divisible by 11
Thus P(k + 1) is true, whenever P(k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n N.
Q21. x2n y2n is divisible by x + y
Solution: Let the given statement be P(n), so
P(n) : x2n – y2n is divisible by x + y
Putting n = 1 we have,
x2n – y2n = x2 - y2 = (x + y) (x - y)
P(n) is true for n = 1
Assume that P(k) is also true for some positive integer k or
x2k – y2k is divisible by (x + y)
So, x2k – y2k = m( x + y)
Or x2k = m( x + y) + y2k …………. (1)
We shall prove that P(k + 1) is true whenever P(k) is true
replacing k by k + 1 we have
x2k + 2 – y2k + 2
= x2k . x2 – y2k .y2
Putting the value of x2k from (1)
= {m( x + y) + y2k} x2 – y2k .y2
= m( x + y) x2 + y2k. x2 – y2k .y2
= m( x + y) x2 + y2k (x2 – y2)
= m( x + y) x2 + y2k (x + y) ( x - y)
= ( x + y) [mx2 + y2k ( x - y)]
x2n – y2n is divisible by x + y
Thus P(k + 1) is true, whenever P(k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n N.
Q22. 32n+2 – 8n – 9 is divisible by 8
Solution: Let the given statement be P(n), so
P(n) : 32n+2 – 8n – 9 is divisible by 8
Putting n =1
P(1) : 32×1+2 – 8 × 1 – 9 = 81 - 17 = 64 = 8 × 8
Which is divisible by 8
P(1) is true
Assume that P(k) is also true for some positive integer k
32k + 2 – 8k – 9
32k + 2 – 8k – 9 is divisible by 8
32k + 2 – 8k – 9 = 8m
Or 32k + 2 = 8m + 8k + 9 ……………. (1)
We shall prove that P(k + 1) is true whenever P(k) is true
replacing k by k + 1 we have
32k + 4 – 8k 8 – 9
= 32k + 4 – 8k 17
= 32k + 2 × 32 – 8k 17
= (8m + 8k + 9)× 9 – 8k 17
= 72m + 72k + 81 – 8k 17
= 72m + 64k + 64
= 8(9m + 8k + 8)
∴ 32n+2 – 8n – 9 is divisible by 8
Thus P(k + 1) is true, whenever P(k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n N.
Q23. 41n – 14n is a multiple of 27.
Solution: Let the given statement be P(n), so
P(n) : 41n – 14n is a multiple of 27
Putting n = 1
P(1): 41n – 14n = 41 – 14 = 27
P(1) is true
Assume that P(k) is also true for some positive integer k
41k – 14k = 27
41k = 27 + 14k ………… (1)
We shall prove that P(k + 1) is true whenever P(k) is true
replacing k by k + 1 we have
41k + 1 – 14k + 1
= 41k . 41 – 14k . 14
= (27 + 14k) 41 – 14k . 14
= 27 . 41 + 14k .41 – 14k . 14
= 27 . 41 + 14k (41 – 14 )
= 27 . 41 + 14k . 27
= 27 ( 41 + 14k )
41n – 14n is a multiple of 27
Thus P(k + 1) is true, whenever P(k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n N.
Q24. (2n + 7) < (n + 3)2
Solution: Let the statement be p(n) so,
p(n) : (2n + 7) < (n + 3)2
=> p(1) : (2 × 1 + 7) < (1 + 3)2
=> 9 < 42
=> 9 < 16
Therefore, p(1) is true so Assume that p(k) is also true for some integer k.
(2k + 7) < (k + 3)2 ......... (i)
Now we shall prove for p(k + 1)
2(k +1) + 7 < (k + 1 + 3)2
2k + 2 + 7 < (k + 4)2 ........ (ii)
We have from (i)
(2k + 7) < (k + 3)2
Adding 2 both sides
=> 2k + 7 + 2 < (k + 3)2 + 2
=> 2k + 7 + 2 < k2 + 6k + 9 + 2
=> 2k + 7 + 2 < k2 + 6k + 9 + 2
=> 2k + 7 + 2 < k2 + 6k + 11
Now, k2 + 6k + 11 < (k + 4)2 from (ii)
=> 2k + 7 + 2 < k+ 6k + 11 < k2 + 8k + 16
=> 2k + 2 + 7 < k2 + 8k + 16
=> 2(k + 1) + 7 < (k + 4)2
=> 2(k + 1) + 7 < (k + 1 + 3)2
Thus P(+ 1) is true, whenever P(k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n ∈ N.
Chapter Contents: |
# Lesson: Lesson 5: Calculating Missing Angles in Parallel Lines Cut By a Transversal
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### Lesson Objective
SWBAT Calculate Angle Measures in Parallel Lines Cut By A Transversal Using Algebra
### Lesson Plan
Opening
Review Angle Relationships from Lesson 4. Have students silently complete the table for Alternate Interior, Alternate Exterior, Corresponding, and Vertical.
Circulate to check students’ progress and accuracy, then have student volunteers fill in the table.
Have students complete the rest of the page, then go over it.
Example 1: Solve for x in terms of Parallel Lines Cut by Transversals
1. Identify the angle relationship: Supplementary
(corresponding, supplementary, alternate interior, or alternate exterior)
This is a great opportunity to review obtuse/acute. Some questions to ask: Which angle is obtuse? Which angle is acute? Since one is acute and one is obtuse can they be congruent? Why or why not? Since they are not congruent, what does that mean? How do you know they add up to 180? How could you prove that using angle relationships?
1. That means the angles…add up to 180 degrees
2. Set up the equation where angles add up to 180.
3. Solve the equation for x
4. Check by substitution
5. Have students complete You Try 1
Example 2: Find the Measure of Angles in Parallel Lines Cut By A Transversal
1. Identify the angle relationship: Alternate Exterior to solve for x.
(corresponding, supplementary, alternate interior, or alternate exterior)
1. That means the angles…are equal to each other
2. Set up the equation where angles are equal to each other
3. Solve the equation for x
4. Plug x back into the angle expression to find the angle measure
5. Write in on the diagram and explicitly model as “best practice” for students
1. Identify the angle relationship: supplementary to solve for x.
(corresponding, supplementary, alternate interior, or alternate exterior)
1. That means the angles…add up to 180
2. Set up the equation where angles add up to 180
3. Subtract to find angle HFE
4. Ask students, what’s the difference between this problem and Example 1? (One you solve for X and not plug it back in.)
5. Have students complete You Try 2
Common Blunders:
1. Students identify the incorrect relationship
2. Students mix up congruent and supplementary
3. Students mix up alternate exterior and alternate interior
4. Students interchange the words complementary and congruent
5. Students solve for X and stop
6. Students plug in for the angle measure when they should only solve for X
Student Work Analysis
Have students apply their knowledge to identify the student mistake (should be corresponding angles so equations are equal to each other). Circulate, asking guiding questions (as seen above). Discuss the students’ mistakes. Have students solve the problem correctly in the box.
Closing
Have students share out and summarize what they learned today.
Assessment
Have students complete the Exit Ticket
Reflection:
What works: Students learn to set up equations effectively and solve for x, as well as take it to the next level by solving for the missing angle. Also, several diagrams are not drawn to scale which invite an interesting discussion.
What didn't work:
Although there are some explicit scaffolds to help students through their thinking, they had a difficult time internalizing the guiding questions. I would add a section where students have to write out/articulate the questions they should ask themselves – almost like a student think-aloud. The metacognitive piece is crucial as my students were not asking themselves “What makes logical sense? What’s the next piece of information you need and how are you going to find it?”
### Lesson Resources
Unit 5 Lesson 5 Calculating Missing Angles Parallel Lines Cut By a Transversal HW.docx 1,130 Unit 5 Lesson 5 Calculating Missing Angles Parallel Lines Cut By a Transversal.docx 1,365 |
# Point-Slope Form
A line is in point-slope form if it looks like
yy1 = m(xx1)
where y1, x1, and m are real numbers. Here (x1, y1) is a fixed point on the line, and m is the slope of the line. In fact, (x1, y1) is so fixed that it's never going to birth a litter. #petjokes
To graph an equation given in point-slope form, it's often easiest to rewrite the equation in slope-intercept form.
### Sample Problem
Graph the equation y – 3 = 4(x – 0.5).
First we add 3 to each side:
y = 4(x – 0.5) + 3
Then simplify to get:
y = 4x + 1
From here, we can graph the equation using the y-intercept and the slope:
Point-slope form is most useful for finding the equation of a line when you're given either a graph or two points on the line. By the way, when you're given a graph, say "thank you" and don't ask for any more. You don't want to look a gift graph in the mouth.
### Sample Problem
Find the equation of the line shown below.
First we need to pick a point (x1, y1). Let's take a point with nice, even integer coordinates. Yes, 14,838 and 372,410 are even numbers, but we can do better. Let (x1, y1) be the point (0, 1), so x1 = 0 and y1 = 1.
Now we need to find the slope, m, of the line. Pick another point on the line and look at the rise and run between the two points. Don't look at anything else if you can help it; this slope is a little self-conscious.
We can conclude that .
To write the equation for the line, we use the blueprint yy1 = m(xx1) and plug in the values x1 = 0, y1 = 1, and .
Rearrange that bad boy to get:
Here's a fun trick (and yeah, we're using "fun" very, very loosely): if we rearrange the point-slope equation yy1 = m(xx1), we find:
If we fix a point (x1, y1) on the line, then for any other point (x, y) on the line we can think of yy1 as the rise and xx1 as the run. We know how much you love your visual aids, and we would never dream of depriving you of them, so here you go:
Since m is the slope of the line, saying is really just saying , which we know is true. And just like that, we've got a handy new formula for finding the slope. |
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# 5.6 The cosine function
## 5.6 The cosine function (EMBH3)
### Revision (EMBH4)
#### Functions of the form $$y = \cos \theta$$ for $$\text{0}\text{°} \leq \theta \leq \text{360}\text{°}$$
• The period is $$\text{360}\text{°}$$ and the amplitude is $$\text{1}$$.
• Domain: $$[\text{0}\text{°};\text{360}\text{°}]$$
For $$y = \cos \theta$$, the domain is $$\{ \theta: \theta \in \mathbb{R} \}$$, however in this case, the domain has been restricted to the interval $$\text{0}\text{°} \leq \theta \leq \text{360}\text{°}$$.
• Range: $$\left[-1;1\right]$$
• $$x$$-intercepts: $$\left(\text{90}\text{°};0\right)$$, $$\left(\text{270}\text{°};0\right)$$
• $$y$$-intercept: $$\left(\text{0}\text{°};1\right)$$
• Maximum turning points: $$\left(\text{0}\text{°};1\right)$$, $$\left(\text{360}\text{°};1\right)$$
• Minimum turning point: $$\left(\text{180}\text{°};-1\right)$$
#### Functions of the form $$y = a \cos \theta + q$$
Cosine functions of the general form $$y = a \cos \theta + q$$, where $$a$$ and $$q$$ are constants.
The effects of $$a$$ and $$q$$ on $$f(\theta) = a \cos \theta + q$$:
• The effect of $$q$$ on vertical shift
• For $$q>0$$, $$f(\theta)$$ is shifted vertically upwards by $$q$$ units.
• For $$q<0$$, $$f(\theta)$$ is shifted vertically downwards by $$q$$ units.
• The effect of $$a$$ on shape
• For $$a>1$$, the amplitude of $$f(\theta)$$ increases.
• For $$0<a<1$$, the amplitude of $$f(\theta)$$ decreases.
• For $$a<0$$, there is a reflection about the $$x$$-axis.
• For $$-1 < a < 0$$, there is a reflection about the $$x$$-axis and the amplitude decreases.
• For $$a < -1$$, there is a reflection about the $$x$$-axis and the amplitude increases.
temp text
## Revision
Textbook Exercise 5.24
On separate axes, accurately draw each of the following functions for $$\text{0}\text{°} \leq \theta \leq \text{360}\text{°}$$:
• Use tables of values if necessary.
• Use graph paper if available.
For each function in the previous problem determine the following:
• Period
• Amplitude
• Domain and range
• $$x$$- and $$y$$-intercepts
• Maximum and minimum turning points
$$y_1 = \cos \theta$$
$$y_2 = - 3 \cos \theta$$
$$y_3 = \cos \theta + 2$$
$$y_4 = \frac{1}{2} \cos \theta - 1$$
### Functions of the form $$y=\cos (k\theta)$$ (EMBH5)
We now consider cosine functions of the form $$y = \cos k\theta$$ and the effects of parameter $$k$$.
## The effects of $$k$$ on a cosine graph
1. Complete the following table for $$y_1 = \cos \theta$$ for $$-\text{360}\text{°} \leq \theta \leq \text{360}\text{°}$$:
θ $$-\text{360}$$$$\text{°}$$ $$-\text{300}$$$$\text{°}$$ $$-\text{240}$$$$\text{°}$$ $$-\text{180}$$$$\text{°}$$ $$-\text{120}$$$$\text{°}$$ $$-\text{60}$$$$\text{°}$$ $$\text{0}$$$$\text{°}$$ $$\cos \theta$$ θ $$\text{60}$$$$\text{°}$$ $$\text{120}$$$$\text{°}$$ $$\text{180}$$$$\text{°}$$ $$\text{240}$$$$\text{°}$$ $$\text{300}$$$$\text{°}$$ $$\text{360}$$$$\text{°}$$ $$\cos \theta$$
2. Use the table of values to plot the graph of $$y_1 = \cos \theta$$ for $$-\text{360}\text{°} \leq \theta \leq \text{360}\text{°}$$.
3. On the same system of axes, plot the following graphs:
1. $$y_2 = \cos (-\theta)$$
2. $$y_3 = \cos 3\theta$$
3. $$y_4 = \cos \frac{3\theta}{4}$$
4. Use your sketches of the functions above to complete the following table:
$$y_1$$ $$y_2$$ $$y_3$$ $$y_4$$ period amplitude domain range maximum turning points minimum turning points $$y$$-intercept(s) $$x$$-intercept(s) effect of $$k$$
5. What do you notice about $$y_1 = \cos \theta$$ and $$y_2 = \cos (-\theta)$$?
6. Is $$\cos (-\theta) = -\cos \theta$$ a true statement? Explain your answer.
7. Can you deduce a formula for determining the period of $$y = \cos k\theta$$?
The effect of the parameter $$k$$ on $$y = \cos k\theta$$
The value of $$k$$ affects the period of the cosine function.
• For $$k > 0$$:
For $$k > 1$$, the period of the cosine function decreases.
For $$0 < k < 1$$, the period of the cosine function increases.
• For $$k < 0$$:
For $$-1 < k < 0$$, the period increases.
For $$k < -1$$, the period decreases.
Negative angles: $\cos (-\theta) = \cos \theta$ Notice that for negative values of $$\theta$$, the graph is not reflected about the $$x$$-axis.
Calculating the period:
To determine the period of $$y = \cos k\theta$$ we use, $\text{Period } = \frac{\text{360}\text{°}}{|k|}$ where $$|k|$$ is the absolute value of $$k$$.
$$0 < k < 1$$ $$-1 < k < 0$$ $$k > 1$$ $$k < -1$$
## Worked example 22: Cosine function
1. Sketch the following functions on the same set of axes for $$-\text{180}\text{°} \leq \theta \leq \text{180}\text{°}$$.
1. $$y_1 = \cos \theta$$
2. $$y_2 = \cos \frac{\theta}{2}$$
2. For each function determine the following:
1. Period
2. Amplitude
3. Domain and range
4. $$x$$- and $$y$$-intercepts
5. Maximum and minimum turning points
### Examine the equations of the form $$y = \cos k\theta$$
Notice that for $$y_2 = \cos \frac{\theta}{2}$$, $$k < 1$$ therefore the period of the graph increases.
### Complete a table of values
θ $$-\text{180}$$$$\text{°}$$ $$-\text{135}$$$$\text{°}$$ $$-\text{90}$$$$\text{°}$$ $$-\text{45}$$$$\text{°}$$ $$\text{0}$$$$\text{°}$$ $$\text{45}$$$$\text{°}$$ $$\text{90}$$$$\text{°}$$ $$\text{135}$$$$\text{°}$$ $$\text{180}$$$$\text{°}$$ $$\cos \theta$$ $$-\text{1}$$ $$-\text{0,71}$$ $$\text{0}$$ $$\text{0,71}$$ $$\text{1}$$ $$\text{0,71}$$ $$\text{0}$$ $$-\text{0,71}$$ $$-\text{1}$$ $$\cos \frac{\theta}{2}$$ $$\text{0}$$ $$\text{0,38}$$ $$\text{0,71}$$ $$\text{0,92}$$ $$\text{1}$$ $$\text{0,92}$$ $$\text{0,71}$$ $$\text{0,38}$$ $$\text{0}$$
### Complete the table
$$y_1 = \cos \theta$$ $$y_2 = \cos \frac{\theta}{2}$$ period $$\text{360}\text{°}$$ $$\text{720}\text{°}$$ amplitude $$\text{1}$$ $$\text{1}$$ domain $$[-\text{180}\text{°};\text{180}\text{°}]$$ $$[-\text{180}\text{°};\text{180}\text{°}]$$ range $$[-1;1]$$ $$[0;1]$$ maximum turning points $$(\text{0}\text{°};1)$$ $$(\text{0}\text{°};1)$$ minimum turning points $$(-\text{180}\text{°};-1) \text{ and } (\text{180}\text{°};-1)$$ none $$y$$-intercept(s) $$(\text{0}\text{°};1)$$ $$(\text{0}\text{°};1)$$ $$x$$-intercept(s) $$(-\text{90}\text{°};0) \text{ and } (\text{90}\text{°};0)$$ $$(-\text{180}\text{°};0) \text{ and } (\text{180}\text{°};0)$$
#### Discovering the characteristics
For functions of the general form: $$f(\theta) = y =\cos k\theta$$:
Domain and range
The domain is $$\{ \theta: \theta \in \mathbb{R} \}$$ because there is no value for $$\theta$$ for which $$f(\theta)$$ is undefined.
The range is $$\{ f(\theta): -1 \leq f(\theta) \leq 1, f(\theta) \in \mathbb{R} \}$$ or $$[-1;1]$$.
Intercepts
The $$x$$-intercepts are determined by letting $$f(\theta) = 0$$ and solving for $$\theta$$.
The $$y$$-intercept is calculated by letting $$\theta = \text{0}\text{°}$$ and solving for $$f(\theta)$$. \begin{align*} y &= \cos k\theta \\ &= \cos \text{0}\text{°} \\ &= 1 \end{align*} This gives the point $$(\text{0}\text{°};1)$$.
temp text
## Cosine functions of the form $$y = \cos k\theta$$
Textbook Exercise 5.25
Sketch the following functions for $$-\text{180}\text{°} \leq \theta \leq \text{180}\text{°}$$. For each graph determine:
• Period
• Amplitude
• Domain and range
• $$x$$- and $$y$$-intercepts
• Maximum and minimum turning points
$$f(\theta) =\cos 2\theta$$
For $$f(\theta) =\cos 2 \theta$$:
\begin{align*} \text{Period: } & \text{180}\text{°} \\ \text{Amplitude: } & 1 \\ \text{Domain: } & [-\text{180}\text{°};\text{180}\text{°}] \\ \text{Range: } & [-1;1] \\ x\text{-intercepts: } & (-\text{135}\text{°};0); (-\text{45}\text{°};0); (\text{45}\text{°};0); (\text{135}\text{°};0) \\ y\text{-intercepts: } & (\text{0}\text{°};1) \\ \text{Max. turning point: } & (-\text{180}\text{°};1); (\text{0}\text{°};1); (\text{180}\text{°};1) \\ \text{Min. turning point: } & (-\text{90}\text{°};-1); (\text{90}\text{°};-1) \end{align*}
$$g(\theta) =\cos \frac{\theta}{3}$$
For $$g(\theta) =\cos \frac{\theta}{3}$$:
\begin{align*} \text{Period: } & \text{1 080}\text{°} \\ \text{Amplitude: } & 1 \\ \text{Domain: } & [-\text{180}\text{°};\text{180}\text{°}] \\ \text{Range: } & [\frac{1}{2};1] \\ x\text{-intercepts: } & \text{ none } \\ \text{Max. turning point: } & (\text{0}\text{°};1) \\ \text{Min. turning point: } & \text{ none } \end{align*}
$$h(\theta) =\cos (-2\theta)$$
For $$h(\theta) =\cos (-2\theta)$$:
\begin{align*} \text{Period: } & \text{180}\text{°} \\ \text{Amplitude: } & 1 \\ \text{Domain: } & [-\text{180}\text{°};\text{180}\text{°}] \\ \text{Range: } & [-1;1] \\ x\text{-intercepts: } & (-\text{135}\text{°};0); (-\text{45}\text{°};0); (\text{45}\text{°};0); (\text{135}\text{°};0) \\ y\text{-intercepts: } & (\text{0}\text{°};1) \\ \text{Max. turning point: } & (-\text{180}\text{°};1); (\text{0}\text{°};1); (\text{180}\text{°};1) \\ \text{Min. turning point: } & (-\text{90}\text{°};-1); (\text{90}\text{°};-1) \end{align*}
$$k(\theta) =\cos \frac{3\theta}{4}$$
For $$k(\theta) =\cos \frac{3\theta}{4}$$:
\begin{align*} \text{Period: } & \text{480}\text{°} \\ \text{Amplitude: } & 1 \\ \text{Domain: } & [-\text{180}\text{°};\text{180}\text{°}] \\ \text{Range: } & [-\frac{1}{\sqrt{2}};1] \\ x\text{-intercepts: } & (-\text{120}\text{°};0); (\text{120}\text{°};0) \\ y\text{-intercepts: } & (\text{0}\text{°};1) \\ \text{Max. turning point: } & (\text{0}\text{°};1) \\ \text{Min. turning point: } & \text{ none } \end{align*}
For each graph of the form $$f(\theta) =\cos k\theta$$, determine the value of $$k$$:
\begin{align*} \text{Period } &= \frac{\text{720}\text{°}}{3 \text{ complete waves }} \\ &= \text{240}\text{°} \\ \therefore \frac{\text{360}\text{°}}{k} &= \text{240}\text{°} \\ \therefore k &= \frac{\text{360}\text{°}}{\text{240}\text{°} } \\ &= \frac{3}{2} \end{align*}
\begin{align*} \text{For } y &= \cos \theta \\ 0 &= \cos \text{90}\text{°}\\ \text{So for } A(\text{135}\text{°};0) \qquad \text{90}\text{°} &= k \times \text{135}\text{°}\\ \therefore k &= \frac{\text{90}\text{°}}{\text{135}\text{°} } \\ &= \frac{2}{3} \end{align*}
### Functions of the form $$y=\cos\left(\theta +p\right)$$ (EMBH6)
We now consider cosine functions of the form $$y = \cos(\theta + p)$$ and the effects of parameter $$p$$.
## The effects of $$p$$ on a cosine graph
1. On the same system of axes, plot the following graphs for $$-\text{360}\text{°} \leq \theta \leq \text{360}\text{°}$$:
1. $$y_1 = \cos \theta$$
2. $$y_2 = \cos (\theta - \text{90}\text{°})$$
3. $$y_3 = \cos (\theta - \text{60}\text{°})$$
4. $$y_4 = \cos (\theta + \text{90}\text{°})$$
5. $$y_5 = \cos (\theta + \text{180}\text{°})$$
2. Use your sketches of the functions above to complete the following table:
$$y_1$$ $$y_2$$ $$y_3$$ $$y_4$$ $$y_5$$ period amplitude domain range maximum turning points minimum turning points $$y$$-intercept(s) $$x$$-intercept(s) effect of $$p$$
The effect of the parameter on $$y = \cos(\theta + p)$$
The effect of $$p$$ on the cosine function is a horizontal shift (or phase shift); the entire graph slides to the left or to the right.
• For $$p > 0$$, the graph of the cosine function shifts to the left by $$p$$ degrees.
• For $$p < 0$$, the graph of the cosine function shifts to the right by $$p$$ degrees.
$$p>0$$ $$p<0$$
## Worked example 23: Cosine function
1. Sketch the following functions on the same set of axes for $$-\text{360}\text{°} \leq \theta \leq \text{360}\text{°}$$.
1. $$y_1 = \cos \theta$$
2. $$y_2 = \cos (\theta + \text{30}\text{°})$$
2. For each function determine the following:
1. Period
2. Amplitude
3. Domain and range
4. $$x$$- and $$y$$-intercepts
5. Maximum and minimum turning points
### Examine the equations of the form $$y = \cos (\theta + p)$$
Notice that for $$y_1 = \cos \theta$$ we have $$p = 0$$ (no phase shift) and for $$y_2 = \cos (\theta + \text{30}\text{°})$$, $$p < 0$$ therefore the graph shifts to the left by $$\text{30}\text{°}$$.
### Complete a table of values
θ $$-\text{360}$$$$\text{°}$$ $$-\text{270}$$$$\text{°}$$ $$-\text{180}$$$$\text{°}$$ $$-\text{90}$$$$\text{°}$$ $$\text{0}$$$$\text{°}$$ $$\text{90}$$$$\text{°}$$ $$\text{180}$$$$\text{°}$$ $$\text{270}$$$$\text{°}$$ $$\text{360}$$$$\text{°}$$ $$\cos \theta$$ $$\text{1}$$ $$\text{0}$$ $$-\text{1}$$ $$\text{0}$$ $$\text{1}$$ $$\text{0}$$ $$-\text{1}$$ $$\text{0}$$ $$\text{1}$$ $$\cos(\theta + \text{30}\text{°})$$ $$\text{0,87}$$ $$-\text{0,5}$$ $$-\text{0,87}$$ $$\text{0,5}$$ $$\text{0,87}$$ $$-\text{0,5}$$ $$-\text{0,87}$$ $$\text{0,5}$$ $$\text{0,87}$$
### Complete the table
$$y_1$$ $$y_2$$ period $$\text{360}\text{°}$$ $$\text{360}\text{°}$$ amplitude $$\text{1}$$ $$\text{1}$$ domain $$[-\text{360}\text{°};\text{360}\text{°}]$$ $$[-\text{360}\text{°};\text{360}\text{°}]$$ range $$[-1;1]$$ $$[-1;1]$$ maximum turning points $$(-\text{360}\text{°};1)$$, $$(\text{0}\text{°};1)$$ and $$(\text{360}\text{°};1)$$ $$(-\text{30}\text{°};1)$$ and $$(\text{330}\text{°};1)$$ minimum turning points $$(-\text{180}\text{°};-1)$$ and $$(\text{180}\text{°};-1)$$ $$(-\text{210}\text{°};-1)$$ and $$(\text{150}\text{°};-1)$$ $$y$$-intercept(s) $$(\text{0}\text{°};0)$$ $$(\text{0}\text{°};\text{0,87})$$ $$x$$-intercept(s) $$(-\text{270}\text{°};0)$$, $$(-\text{90}\text{°};0)$$, $$(\text{90}\text{°};0)$$ and $$(\text{270}\text{°};0)$$ $$(-\text{300}\text{°};0)$$, $$(-\text{120}\text{°};0)$$, $$(\text{60}\text{°};0)$$ and $$(\text{240}\text{°};0)$$
#### Discovering the characteristics
For functions of the general form: $$f(\theta) = y =\cos (\theta + p)$$:
Domain and range
The domain is $$\{ \theta: \theta \in \mathbb{R} \}$$ because there is no value for $$\theta$$ for which $$f(\theta)$$ is undefined.
The range is $$\{ f(\theta): -1 \leq f(\theta) \leq 1, f(\theta) \in \mathbb{R} \}$$.
Intercepts
The $$x$$-intercepts are determined by letting $$f(\theta) = 0$$ and solving for $$\theta$$.
The $$y$$-intercept is calculated by letting $$\theta = \text{0}\text{°}$$ and solving for $$f(\theta)$$. \begin{align*} y &= \cos (\theta + p) \\ &= \cos (\text{0}\text{°} + p) \\ &= \cos p \end{align*} This gives the point $$(\text{0}\text{°};\cos p)$$.
temp text
## Cosine functions of the form $$y = \cos (\theta + p)$$
Textbook Exercise 5.26
Sketch the following functions for $$-\text{360}\text{°} \leq \theta \leq \text{360}\text{°}$$.
For each function, determine the following:
• Period
• Amplitude
• Domain and range
• $$x$$- and $$y$$-intercepts
• Maximum and minimum turning points
$$f(\theta) =\cos (\theta + \text{45}\text{°})$$
$$g(\theta) =\cos (\theta - \text{30}\text{°})$$
$$h(\theta) =\cos (\theta + \text{60}\text{°})$$
## Worked example 24: Sketching a cosine graph
Sketch the graph of $$f(\theta) = \cos (\text{180}\text{°} - 3\theta)$$ for $$\text{0}\text{°} \leq \theta \leq \text{360}\text{°}$$.
### Examine the form of the equation
Write the equation in the form $$y = \cos k(\theta + p)$$. \begin{align*} f(\theta) &= \cos (\text{180}\text{°
To draw a graph of the above function, the standard cosine graph, $$y = \cos \theta$$, must be changed in the following ways:
• decrease the period by a factor of $$\text{3}$$
• shift to the right by $$\text{60}\text{°}$$.
### Complete a table of values
θ $$\text{0}$$$$\text{°}$$ $$\text{45}$$$$\text{°}$$ $$\text{90}$$$$\text{°}$$ $$\text{135}$$$$\text{°}$$ $$\text{180}$$$$\text{°}$$ $$\text{225}$$$$\text{°}$$ $$\text{270}$$$$\text{°}$$ $$\text{315}$$$$\text{°}$$ $$\text{360}$$$$\text{°}$$ $$f(\theta)$$ $$-\text{1}$$ $$\text{0,71}$$ $$\text{0}$$ $$-\text{0,71}$$ $$\text{1}$$ $$-\text{0,71}$$ $$\text{0}$$ $$\text{0,71}$$ $$-\text{1}$$
### Plot the points and join with a smooth curve
Period: $$\text{120}$$$$\text{°}$$
Amplitude: $$\text{1}$$
Domain: $$[\text{0}\text{°};\text{360}\text{°}]$$
Range: $$[-1;1]$$
Maximum turning point: $$(\text{60}\text{°};1)$$, $$(\text{180}\text{°};1)$$ and $$(\text{300}\text{°};1)$$
Minimum turning point: $$(\text{0}\text{°}; -1)$$, $$(\text{120}\text{°};-1)$$, $$(\text{240}\text{°};-1)$$ and $$(\text{360}\text{°};-1)$$
$$y$$-intercepts: $$(\text{0}\text{°};-1)$$
$$x$$-intercept: $$(\text{30}\text{°};0)$$, $$(\text{90}\text{°};0)$$, $$(\text{150}\text{°};0)$$, $$(\text{210}\text{°};0)$$, $$(\text{270}\text{°};0)$$ and $$(\text{330}\text{°};0)$$
## Worked example 25: Finding the equation of a cosine graph
Given the graph of $$y = a \cos (k\theta + p)$$, determine the values of $$a$$, $$k$$, $$p$$ and the minimum turning point.
### Determine the value of $$k$$
From the sketch we see that the period of the graph is $$\text{360}\text{°}$$, therefore $$k = 1$$.
$y = a \cos ( \theta + p)$
### Determine the value of $$a$$
From the sketch we see that the maximum turning point is $$(\text{45}\text{°};2)$$, so we know that the amplitude of the graph is $$\text{2}$$ and therefore $$a = 2$$.
$y = 2 \cos ( \theta + p)$
### Determine the value of $$p$$
Compare the given graph with the standard cosine function $$y = \cos \theta$$ and notice the difference in the maximum turning points. We see that the given function has been shifted to the right by $$\text{45}$$$$\text{°}$$, therefore $$p = \text{45}\text{°}$$.
$y = 2 \cos ( \theta - \text{45}\text{°})$
### Determine the minimum turning point
At the minimum turning point, $$y = -2$$:
\begin{align*} y &= 2 \cos ( \theta - \text{45}\text{°}) \\ -2 &= 2 \cos ( \theta - \text{45}\text{°}) \\ -1 &= \cos ( \theta - \text{45}\text{°}) \\ \cos^{-1}(-1) &= \theta - \text{45}\text{°} \\ \text{180}\text{°} &= \theta - \text{45}\text{°} \\ \text{225}\text{°} &= \theta \end{align*}
This gives the point $$(\text{225}\text{°};-2)$$.
## The cosine function
Textbook Exercise 5.27
Sketch the following graphs on separate axes:
$$y = \cos (\theta + \text{15}\text{°})$$ for $$-\text{180}\text{°} \leq \theta \leq \text{180}\text{°}$$
$$f(\theta) = \frac{1}{3} \cos (\theta - \text{60}\text{°})$$ for $$-\text{90}\text{°} \leq \theta \leq \text{90}\text{°}$$
$$y = -2 \cos \theta$$ for $$\text{0}\text{°} \leq \theta \leq \text{360}\text{°}$$
$$y = \cos (\text{30}\text{°} - \theta)$$ for $$-\text{360}\text{°} \leq \theta \leq \text{360}\text{°}$$
\begin{align*} y &= \cos (\text{30}\text{°} - \theta) \\ &= \cos \left( -(\theta - \text{30}\text{°}) \right) \\ &= \cos \left(\theta - \text{30}\text{°} \right) \end{align*}
$$g(\theta) = 1 + \cos (\theta - \text{90}\text{°})$$ for $$\text{0}\text{°} \leq \theta \leq \text{360}\text{°}$$
$$y = \cos (2 \theta + \text{60}\text{°})$$ for $$-\text{360}\text{°} \leq \theta \leq \text{360}\text{°}$$
Two girls are given the following graph:
Audrey decides that the equation for the graph is a cosine function of the form $$y = a \cos \theta$$. Determine the value of $$a$$.
$$a = -1$$
Megan thinks that the equation for the graph is a cosine function of the form $$y = \cos (\theta + p)$$. Determine the value of $$p$$.
$$p = -\text{180}\text{°}$$
What can they conclude?
$$\cos (\theta - \text{180}\text{°}) = -\cos \theta$$ |
# Probability Fraction Calculator
Created by Davide Borchia
Reviewed by Anna Szczepanek, PhD
Last updated: Jan 18, 2024
Following the definition of probability, we can easily calculate probability as a fraction: with our tool, it will be super easy, barely an inconvenience. If you need to calculate the probability as a fraction for multiple events, you are in the right place! Keep reading for a quick explanation of the math behind the calculations, examples, and applications of the fractional representation of probability.
## What is the probability of an event?
The probability of an event is the measure of the frequency with which said event happens out of a total possible amount of outcomes. If you are dealing with coin tosses, for example, you may find out that head is the result in $495$ out of $1000$ tosses.
There are many ways to express probability, but in general, they all stem from its representation as the ratio between the occurrences of a given outcome and the total number of events happening:
$n_{\mathrm{outcome}}:n_{\mathrm{total}}$
We are used to seeing this ratio expressed as a decimal number, the result of the division of the two members, or as a percentage (the same result, multiplied by $100$. There's, however, an additional way to express probability, and it may come in handy in specific situations: in the next section, we will learn how to calculate probability in fraction form.
## Probability as a fraction
To express probability as a fraction, simply write the number of events that resulted in the desired outcome as the numerator of the fraction and the total number of realizations as the denominator.
You can easily calculate the fraction form of probability with the following formula:
$P(\mathrm{A}) = \frac{n_{\mathrm{A}}}{n_{\mathrm{total}}}$
Where:
• $P(\mathrm{A})$ — The probability of the outcome $\mathrm{A}$;
• $n_{\mathrm{A}}$ — The number of times the event had outcome $\mathrm{A}$; and
• $n_{\mathrm{total}}$ — The total number of events from which we consider the selected outcome.
To calculate the probability as a fraction, follow these steps:
1. Find the number of outcomes and the total number of repetitions.
2. Write the number of outcomes as the numerator and the total number of repetitions as the denominator.
3. Calculate the greatest common divisor of these two numbers. Visit Omni's GCF calculator if you need a refresh on the topic!
4. If the GCF is larger than $1$, divide both the numerator and the denominator of the probability fraction by its value: you will obtain the reduced form of the probability.
## How do I calculate probability as a fraction: the case of multiple events.
In case of multiple outcomes of an event (think of a die and its six faces), we can still calculate the probability as a fraction; however, we need to introduce some small modifications to the process!
Say that you are dealing with an event with possible outcomes $\mathrm{A}$, $\mathrm{B}$, and $\mathrm{C}$. Each of the outcomes happened with the following results:
$\begin{split} \mathrm{A}& \rightarrow n_{\mathrm{A}}\\ \mathrm{B}& \rightarrow n_{\mathrm{B}}\\ \mathrm{C}& \rightarrow n_{\mathrm{C}}\\ \end{split}$
If we sum the occurrences, we find the total number of "realizations":
$n_{\mathrm{total}} = n_{\mathrm{A}}+ n_{\mathrm{B}}+ n_{\mathrm{C}}$
Then, knowing that probability can be a fraction, for the first outcome we write the following expressions:
$P(\mathrm{A}) = \frac{n_{\mathrm{A}}}{n_{\mathrm{total}}}$
For the second and third outcomes, we have, respectively:
$P(\mathrm{B}) = \frac{n_{\mathrm{B}}}{n_{\mathrm{total}}}$
And:
$P(\mathrm{C}) = \frac{n_{\mathrm{C}}}{n_{\mathrm{total}}}$
As a fundamental rule, if the considered outcomes span all the realizations of the event, the following rule holds:
$P(\mathrm{A})+P(\mathrm{B})+P(\mathrm{C})=1$
Calculating the probability of multiple events as a fraction is closely related to another way to represent the same quantities! Think about it: if your outcomes span all the possible events, then their fraction will sum to unity. If we compare unity to a full angle, we can represent the partition of the outcomes as sectors in a pie chart. To understand this comparison even better, visit our three related tools:
With a graphical representation, the analogy will be clear!
## FAQ
### Can probability be a fraction?
Yes: since we define probability as the ratio between the number of events that resulted in a given outcome and the total number of events, we can write these two numbers as the numerator and denominator of a fraction. The fractional representation of probability gives us a quick indication of the magnitude of the probability since its value easily compares to the unit fraction 1.
### How do I find probability in fractions?
To calculate probability in fraction form, follow these easy steps:
1. Find the number of events that resulted in the desired outcome. We call this number nA.
2. Find the total number of realizations, nTOT.
3. Define the fractional form of the probability of the event A as:
P(A) = nA/nTOT
4. Calculate the greater common factor of nA and nTOT. If it's different from 1, divide both numbers by the factor, and find the reduced form of the fraction.
### What is the fraction form of the probability of the results of a coin toss?
1/2 for heads and 1/2 for tails. What do these numbers mean? Take the first fraction:
• The 2 at the denominator is the total number of tosses;
• The 1 at the numerator is the number of tosses resulting in heads.
However, getting two tails or two heads is not so unlikely. Try tossing 1000 times. Heads may be the result of, say, 504 tosses. The fraction 504/1000 is similar in value to 1/2. Repeating the event gives a more accurate fractional representation of the probability.
### How do I calculate the fractional form of the probability of multiple events?
To calculate the fractional form of the probability of multiple events:
1. Define the number of events with defined outcomes: nA, nB, nC, and so on.
2. Calculate the number of total events: nTOT = nA + nB + nC + ...
3. Write the desired number of fractions in the form nA/nTOT, nB/nTOT, nC/nTOT, etc.
4. Simplify the fractions, if possible, using the greater common factor between the respective numerator and the denominator.
Davide Borchia
Enter values in each group
Outcome A
Outcome B
Outcome C
Outcome D
Outcome E
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# 5.10: Law of Sines
Difficulty Level: At Grade Created by: CK-12
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Practice Law of Sines
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While working in art class you are trying to design pieces of glass that you will eventually fit together into a sculpture. You are drawing out what you think will be a diagram of one of the pieces. You have a side of length 14 inches, and side of length 17 inches, and an angle next to the 17 in side of 35\begin{align*}35^\circ\end{align*} (not the angle between the 14 in and 17 in pieces, but the one at the other end of the 17 in piece). Your diagram looks like this:
It occurs to you that you could use your knowledge of math to find out if you are going to be able to finish the drawing and make a piece that could actually be built.
Can you figure out how to do this?
By the end of this Concept, you'll know how to apply the Law of Sines to determine the number of possible solutions for a triangle.
### Watch This
James Sousa: The Law of Sines: The Ambiguous Case
### Guidance
In ABC\begin{align*}\triangle ABC \end{align*} below, we know two sides and a non-included angle. Remember that the Law of Sines states: sinAa=sinBb\begin{align*}\frac{\sin A}{a} = \frac{\sin B}{b}\end{align*}. Since we know a,b\begin{align*}a, b\end{align*}, and A\begin{align*}\angle{A}\end{align*}, we can use the Law of Sines to find B\begin{align*}\angle{B}\end{align*}. However, since this is the SSA case, we have to watch out for the Ambiguous case. Since a<b\begin{align*}a < b\end{align*}, we could be faced with situations where either no possible triangles exist, one possible triangle exists, or two possible triangles exist.
To find out how many solutions there are in an ambiguous case, compare the length of a\begin{align*}a\end{align*} to bsinA\begin{align*}b \sin A\end{align*}. If a<bsinA\begin{align*}a < b \sin A\end{align*}, then there are no solutions. If a=bsinA\begin{align*}a = b \sin A\end{align*}, then there is one solution. If a>bsinA\begin{align*}a > b \sin A\end{align*}, then there are two solutions.
#### Example A
Find B\begin{align*}\angle{B}\end{align*}.
Solution: Use the Law of Sines to determine the angle.
sin411223sin4123sin41121.257446472=sinB23=12sinB=sinB=sinB
Since no angle exists with a sine greater than 1, there is no solution to this problem.
We also could have compared a\begin{align*}a\end{align*} and bsinA\begin{align*}b \sin A\end{align*} beforehand to see how many solutions there were to this triangle.
a=12,bsinA=15.1\begin{align*}a = 12, b \sin A = 15.1\end{align*}: since 12<15.1,a<bsinA\begin{align*}12 < 15.1, a < b \sin A\end{align*} which tells us there are no solutions.
#### Example B
In ABC,a=15,b=20\begin{align*}\triangle ABC, a = 15, b = 20\end{align*}, and A=30\begin{align*}\angle{A} = 30^\circ\end{align*}. Find B\begin{align*}\angle{B}\end{align*}.
Solution: Again in this case, a<b\begin{align*}a < b\end{align*} and we know two sides and a non-included angle. By comparing a\begin{align*}a\end{align*} and bsinA\begin{align*}b \sin A\end{align*}, we find that a=15,bsinA=10\begin{align*}a = 15, b \sin A = 10\end{align*}. Since 15>10\begin{align*}15 > 10\end{align*} we know that there will be two solutions to this problem.
sin301520sin3020sin30150.6666667B=sinB20=15sinB=sinB=sinB=41.8
There are two angles less than 180\begin{align*}180^\circ\end{align*} with a sine of 0.6666667, however. We found the first one, 41.8\begin{align*}41.8^\circ\end{align*}, by using the inverse sine function. To find the second one, we will subtract 41.8\begin{align*}41.8^\circ\end{align*} from 180,B=18041.8=138.2\begin{align*}180^\circ, \angle{B} = 180^\circ - 41.8^\circ = 138.2^\circ\end{align*}.
To check to make sure 138.2\begin{align*}138.2^\circ\end{align*} is a solution, we will use the Triangle Sum Theorem to find the third angle. Remember that all three angles must add up to 180\begin{align*}180^\circ\end{align*}.
180(30+41.8)=108.2or180(30+138.2)=11.8
This problem yields two solutions. Either B=41.8\begin{align*}\angle{B} = 41.8^\circ\end{align*} or 138.2\begin{align*}138.2^\circ\end{align*}.
#### Example C
A boat leaves lighthouse A\begin{align*}A\end{align*} and travels 63km. It is spotted from lighthouse B\begin{align*}B\end{align*}, which is 82km away from lighthouse A\begin{align*}A\end{align*}. The boat forms an angle of 65.1\begin{align*}65.1^\circ\end{align*} with both lighthouses. How far is the boat from lighthouse B\begin{align*}B\end{align*}?
Solution: In this problem, we again have the SSA angle case. In order to find the distance from the boat to the lighthouse (a) we will first need to find the measure of A\begin{align*}\angle A\end{align*}. In order to find A\begin{align*}\angle A\end{align*}, we must first use the Law of Sines to find B\begin{align*}\angle B\end{align*}. Since c>b\begin{align*}c > b\end{align*}, this situation will yield exactly one answer for the measure of B\begin{align*}\angle B\end{align*}.
sin65.18263sin65.1820.6969B=sinB63=sinBsinB=44.2
Now that we know the measure of B\begin{align*}\angle B\end{align*}, we can find the measure of angle A,A=18065.144.2=70.7\begin{align*}A, \angle{A} = 180^\circ - 65.1^\circ - 44.2^\circ = 70.7^\circ\end{align*}. Finally, we can use A\begin{align*}\angle{A}\end{align*} to find side a\begin{align*}a\end{align*}.
sin65.18282sin70.7sin65.1a=sin70.7a=a=85.3
The boat is approximately 85.3 km away from lighthouse B\begin{align*}B\end{align*}.
### Guided Practice
1. Prove using the Law of Sines: acc=sinAsinCsinC\begin{align*}\frac{a-c}{c} = \frac{\sin A - \sin C}{\sin C}\end{align*}
2. Find all possible measures of angle B\begin{align*}B\end{align*} if any exist for the following triangle values: A=32.5,a=26,b=37\begin{align*}A = 32.5^\circ, a = 26, b = 37\end{align*}
3. Find all possible measures of angle B\begin{align*}B\end{align*} if any exist for the following triangle values: A=42.3,a=16,b=26\begin{align*}A = 42.3^\circ, a = 16, b = 26\end{align*}
Solutions:
1.
sinAacsinAcsinAcsinCc(sinAsinC)sinAsinCsinC=sinCc=asinC=asinCcsinC=sinC(ac)=acc
2. sin32.526=sinB37B=49.9\begin{align*}\frac{\sin 32.5^\circ}{26} = \frac{\sin B}{37} \rightarrow B = 49.9^\circ\end{align*} or 18049.9=130.1\begin{align*}180^\circ - 49.9^\circ = 130.1^\circ\end{align*}
3. no solution
### Concept Problem Solution
A drawing of this situation looks like this:
You can start by using the Law of Sines:
sinAa=sinBb\begin{align*}\frac{\sin A}{a} = \frac{\sin B}{b}\end{align*}
and substitute known values:
sin3514=sinB17\begin{align*}\frac{\sin 35}{14} = \frac{\sin B}{17}\end{align*}
Then solving for sinB\begin{align*}\sin B\end{align*}:
sinB=17sin3514\begin{align*}\sin B = \frac{17\sin 35^\circ}{14}\end{align*}
And so
B44.15\begin{align*}B \approx 44.15^\circ\end{align*}
Since the interior angles of any triangle add up to 180\begin{align*}180^\circ\end{align*}, we can find C\begin{align*}\angle C \end{align*}:
C=1803544.15C=100.85
This information can be used again in the Law of Sines:
sinAa=sinCcsin3514=sin100.86cc=14sin100.86sin35c=13.75.5735c=23.976
### Explore More
Find all possible measures of angle \begin{align*}B\end{align*} if any exist for each of the following triangle values.
1. \begin{align*}A = 30^\circ, a = 13, b = 15\end{align*}
2. \begin{align*}A = 42^\circ, a = 21, b = 12\end{align*}
3. \begin{align*}A = 22^\circ, a = 36, b = 37\end{align*}
4. \begin{align*}A = 87^\circ, a = 14, b = 12\end{align*}
5. \begin{align*}A = 31^\circ, a = 25, b = 44\end{align*}
6. \begin{align*}A = 59^\circ, a = 37, b = 41\end{align*}
7. \begin{align*}A = 81^\circ, a = 22, b = 20\end{align*}
8. \begin{align*}A = 95^\circ, a = 31, b = 34\end{align*}
9. \begin{align*}A = 112^\circ, a = 12, b = 15\end{align*}
10. \begin{align*}A = 78^\circ, a = 20, b = 16\end{align*}
11. In \begin{align*}\triangle ABC\end{align*}, a=10 and \begin{align*}m\angle B=39^\circ\end{align*}. What's a possible value for b that would produce two triangles?
12. In \begin{align*}\triangle ABC\end{align*}, a=15 and \begin{align*}m\angle B=67^\circ\end{align*}. What's a possible value for b that would produce no triangles?
13. In \begin{align*}\triangle ABC\end{align*}, a=21 and \begin{align*}m\angle B=99^\circ\end{align*}. What's a possible value for b that would produce one triangle?
14. Bill and Connie are each leaving for school. Connie's house is 4 miles due east of Bill's house. Bill can see the school in the direction \begin{align*}40^\circ\end{align*} east of north. Connie can see the school on a line \begin{align*}51^\circ\end{align*} west of north. What is the straight line distance of each person from the school?
15. Rochelle and Rose are each looking at a hot air balloon. They are standing 2 miles apart. The angle of elevation for Rochelle is \begin{align*}30^\circ\end{align*} and the angle of elevation for Rose is \begin{align*}34^\circ\end{align*}. How high off the ground is the balloon?
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 5.10.
### Vocabulary Language: English
ambiguous
ambiguous
Ambiguous means that the given information is not specific. In the context of Geometry or Trigonometry, it means that the given data may not uniquely identify one shape.
law of cosines
law of cosines
The law of cosines is a rule relating the sides of a triangle to the cosine of one of its angles. The law of cosines states that $c^2=a^2+b^2-2ab\cos C$, where $C$ is the angle across from side $c$.
law of sines
law of sines
The law of sines is a rule applied to triangles stating that the ratio of the sine of an angle to the side opposite that angle is equal to the ratio of the sine of another angle in the triangle to the side opposite that angle.
SSA
SSA
SSA means side, side, angle and refers to the fact that two sides and the non-included angle of a triangle are known in a problem.
## Date Created:
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# Adding Fractions on a Number Line
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## Adding Fractions on a Number Line
When we learn how to add fractions, we can use visual methods to understand the process better. In this learning text we are going to use number lines as a visual explanation to get a better understanding.
## Steps to Adding Fractions on a Number Line
We know that a fraction has a top number (numerator) and a bottom number (denominator). Adding fractions is different from adding whole numbers. Let’s look at the steps to take if you want to add two fractions on a number line.
• Firstly, check if the fractions have common or like denominators.
• Secondly, divide the number line (between 0 and 1) into equal parts and label each part; the denominator is the indicator in how many parts the number line will be divided between whole numbers. Then look at the first fraction and circle or highlight it on the number line.
• Finally jump from the highlighted fraction to the right as many times as shown by the numerator of the second fraction. Remember to simplify your answer if possible!
## Adding Fractions on a Number Line – Examples
Let’s look at the examples below and follow the process of adding two fractions on a number line. We are adding here two fractions with a common denominator of eight: $\frac{1}{8}$ and $\frac{5}{8}$. In order to add two fractions using the number line, we divide the number line between zero and one into equal parts and label each part. Then we must find the first fraction on our number line and then jump to the right as many times as the numerator of the fraction we are adding shows. Our added fraction has a numerator of five, so we must jump five times forward. We have landed on $\frac{6}{8}$, so $\frac{1}{8}$ add $\frac{5}{8}$ is $\frac{6}{8}$. Now we can simplify our answer if possible. $\frac{6}{8}$ we can simplify to $\frac{3}{4}$ by dividing the numerator and denominator by a common factor which is two.
Let’s look at another example of adding fractions with a numberline.
This time we have $\frac{2}{6}$ and $\frac{3}{6}$. Both fractions share the same denominator, which is six. We are going to repeat the same process as above. We divide the number line between zero and one into equal parts and label each part. Then we must find the first fraction on our number line and then jump to the right as many times as the numerator of the fraction we are adding shows. Our added fraction has a numerator of three, so we must jump three times forward. We have landed on $\frac{5}{6}$, so $\frac{2}{6}$ add $\frac{3}{6}$ is $\frac{5}{6}$. Now we can simplify our answer if possible. This time we cannot simplify further, so our answer is $\frac{5}{6}$.
## Adding Fractions on a Number Line - Further Practice
Today we learned about adding fractions using a number line. Let’s look at the steps below for a quick review.
Step # What to do
1. Check that the fractions have the same denominators.
2. Divide the number line into equal parts between
0 and 1 as shown by the denominators.
3. Find your first fraction on your number line
and circle or highlight it.
4. Jump forward to the right as many times as
shown by the numerator to find the sum.
5. Simplify your answer if possible
To test your knowledge on adding and subtracting fractions on a number line, have a look at our practice problems, videos and worksheets.
## Frequently Asked Questions on Adding Fractions on a Number Line
What are numerators and denominators?
What are common denominators?
How can we find common denominators?
How do you add fractions on a number line with common denominators?
### TranscriptAdding Fractions on a Number Line
"While I fill the tank, keep your eye on the fuel gauge!" "You got it partner, I'll calculate the total fuel we have!" Let's help Tank calculate the total amount of fuel in the submarine by "Adding Fractions on a Number Line." We can use a number line like this to help us when adding fractions. To add fractions on a number line, first, check that the fractions have the same denominator. One-eighth and five-eighths have the same denominator because the number on the bottom part of each fraction is the same. Next, divide the number line into equal parts, using the denominator to determine the number of parts. Eight is the denominator, so make eight equal parts between zero and one and label them like this. Now find one-eighth on the number line which is here. Then identify the numerator of the fraction we are adding which is five. Jump forward five parts from one-eighth. We land on six-eighths. One-eighth plus five-eighths is six-eighths. Finally, simplify the answer if possible. To simplify fractions, find a common factor for the numerator and denominator that both can be divided into. Six-eighths can be simplified by dividing the numerator and denominator by two making the fraction three-quarters. Now we have looked at the steps needed to add fractions on a number line, let's help calculate how much petrol Axel and Tank have in their submarine! The submarine had two-sixths of petrol left in the tank and Axel added three-sixths to the submarine. With the number line ready, what is the first step? First, check that the fractions have the same denominator. Since both fractions have a six as the bottom part of the fraction, we know that they have the same denominator. What is the next step? Divide the number line into equal parts between zero and one as shown by the denominator, which is six, and label each part on the number line. What should we do next? Find the first fraction, two-sixths which is here. How do we find the sum? We identify the numerator of the fraction we are adding which is three and jump forward three parts from two-sixths. The sum of two-sixths plus three-sixths is five-sixths. Can five-sixths be simplified? Five-sixths cannot be simplified as one is the only factor that goes into both five and six. So we leave the answer as five-sixths. While Axel pays for the petrol, let's review! Remember, when adding fractions on a number line: First, check that the fractions have the same denominator. Next, divide the number line into equal parts between whole numbers as shown by the denominator. Then, locate the first fraction on the number line. Finally, jump forward the number of parts as shown by the numerator of the second fraction to find the sum. Remember to simplify the fraction if you can. "Alright, are you ready to hit the road, Tank?" "I was born ready, partner."
## Adding Fractions on a Number Line exercise
Would you like to apply the knowledge you’ve learnt? You can review and practice it with the tasks for the video Adding Fractions on a Number Line.
• ### What are the steps to add fractions?
Hints
Before you start adding the fractions, what do you need to check is the same?
Once you have divided your number line into equal parts, what do you need to locate?
Solution
First, check that the fractions you are adding have the same denominator.
Next, divide the number line into equal parts as shown by the denominators.
Then, locate the first fraction on the number line.
Finally, count forward the number of parts, as shown by the numerator on the second fraction.
• ### How much petrol is in the tank?
Hints
Each interval on the number line goes up in steps of $\frac1 6$. How many jumps will be needed to add $\frac3 6$?
There was already $\frac1 6$ in the tank, so start your jumps from here.
Solution
• As we are adding $\frac1 6$ + $\frac3 6$, we start at $\frac1 6$.
• Next, look at the numerator of the fraction we are adding; in $\frac3 6$ the numerator is 3 so we make 3 jumps.
• This gets us to $\frac4 6$.
• So $\frac1 6$ + $\frac3 6$ = $\frac4 6$
• ### Which number lines show the correct way of adding the fractions?
Hints
The number line should be divided into equal parts based on the numerator. What is the numerator in the fractions that Axel and Tank are adding here?
The friends added $\frac1 9$ and $\frac6 9$. There are two ways to add these fractions, depending on which order they are added.
To add $\frac1 9$ and $\frac6 9$, the friends could start at $\frac1 9$ and make 6 jumps, or they could start at $\frac6 9$ and make 1 jump.
Solution
• There are two correct options to add $\frac1 9$ + $\frac6 9$.
• Both correct options have the number line divided into 9 equal parts because the numerators in $\frac1 9$ + $\frac6 9$ are 9.
• To solve starting with the smaller fraction: $\frac1 9$ + $\frac6 9$, start at $\frac1 9$ and jump forward 6.
• To solve starting with the larger fraction: $\frac6 9$ + $\frac1 9$, start at $\frac6 9$ and jump forward 1.
• ### Practise adding fractions.
Hints
To add the fractions on a number line, first partition the number line to the number of parts that is in the denominator.
Find the first fraction on the number line, then count forward by the numerator of the second fraction.
Can you simplify your answer by dividing the numerator and denominator by the same factor?
Solution
• $\frac2 8$ + $\frac2 8$ = $\frac1 2$.
Start on $\frac2 8$, count forward by two which takes you to $\frac4 8$. $\frac4 8$ can be simplified to $\frac1 2$ by dividing both the numerator (4) and the denominator (8) by 4.
• $\frac3 6$ + $\frac1 6$ = $\frac2 3$.
Start on $\frac3 6$, count forward by one which takes you to $\frac4 6$. $\frac4 6$ can be simplified to $\frac2 3$ by dividing both the numerator (4) and the denominator (6) by 2.
• $\frac1 7$ + $\frac3 7$ = $\frac4 7$.
Start on $\frac1 7$, count forward by three which takes you to $\frac4 7$. This cannot be simplified any further.
• $\frac3 5$ + $\frac2 5$ = 1.
Start on $\frac3 5$, count forward by two which takes you to $\frac5 5$. $\frac5 5$ can be simplified to 1 by dividing both the numerator (5) and the denominator (5) by 5.
• ### Add the fractions on the number line.
Hints
Start by locating the first fraction in the equation on the number line.
How many parts do you need to jump forward?
The numerator in the second fraction ($\frac4 7$) is 4, so jump forward 4 parts.
Solution
• Start at $\frac2 7$
• As we are adding $\frac4 7$, look at the numerator of that fraction.
• The numerator of $\frac4 7$ is 4, so we make 4 jumps forward.
• This gets us to $\frac6 7$.
• So $\frac2 7$ + $\frac4 7$ = $\frac6 7$.
• ### Adding and simplifying fractions.
Hints
To simplify a fraction, divide the numerator and denominator by the same factor. In this example $\frac{4}{10}$ is simplified to $\frac2 5$ by dividing both by 2.
Sometimes, it may be a fraction in the question that has already been simplified and needs expanding. For example, $\frac1 3$ can be expanded by multiplying both the numerator and denominator by 2 to get $\frac2 6$.
Solution
1) This answer was correct. $\frac3 8$ + $\frac3 8$ = $\frac6 8$. Divide numerator and denominator by 2 to get $\frac3 4$.
2) This answer was correct. $\frac4 9$ + $\frac1 3$ = $\frac7 9$. First expand $\frac1 3$ by multiplying the numerator and denominator by 3 to get $\frac3 9$. $\frac4 9$ + $\frac3 9$ = $\frac7 9$.
3) This answer was incorrect. $\frac2 6$ + $\frac2 6$ = $\frac4 6$. Divide numerator and denominator by 2 to get $\frac2 3$.
4) This answer was incorrect. $\frac{4}{10}$ + $\frac{4}{10}$ = $\frac{8}{10}$. Divide numerator and denominator by 2 to get $\frac4 5$.
5) This answer was correct. $\frac{3}{12}$ + $\frac{5}{12}$ = $\frac{8}{12}$. Divide numerator and denominator by 4 to get $\frac2 3$.
6) This answer was correct. $\frac{1}{16}$ + $\frac{1}{16}$ + $\frac{2}{16}$ = $\frac{4}{16}$. Divide numerator and denominator by 4 to get $\frac1 4$. |
# Surface Area of a Cylinder
## Surface Area of a Cylinder Lesson
### Cylinder Surface Area Formula
The formula for surface area of a cylinder is given as:
SA = 2πrh + 2πr2
Where SA is the surface area, r is the radius of the base circle, and h is the height of the cylinder.
INTRODUCING
### Surface Area of a Cylinder Example Problems
Let's go through a couple of example problems together to practice finding the surface area of a cylinder.
#### Example Problem 1
What is the surface area of a cylinder with a diameter of 20 and a height of 20?
Solution:
1. Let's convert diameter to radius. D = 2r, 20 = 2r, so r = 10.
2. Now let's plug the radius and height into the surface area formula.
3. SA = 2πrh + 2πr2
4. SA = 2π(10)(20) + 2π(102) = 2π(200) + 2π(100)
5. SA = 1,884.956
6. The surface area is 1,884.956.
#### Example Problem 2
A cylinder is measured to have a surface area of 24 square meters. It's height and radius are equal. What is the height of the cylinder?
Solution:
1. Let's set up relations between height and radius that we can substitute into the formula.
2. h = r
3. rh = h2
4. r2 = h2
5. Now let's plug what we know into the surface area formula and solve for h.
6. SA = 2πrh + 2πr2
7. 24 = 2πh2 + 2πh2
8. 24 = 4πh2
9. 1.910 = h2
10. h = 1.382
11. The height of the cylinder is 1.382 meters.
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# 6.05 Extension: Properties of logarithms
Lesson
## Laws of exponents and laws of logarithms
Recall that the logarithmic and exponential functions “undo” each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove.
$\log_b1=0$logb1=0
$\log_bb=1$logbb=1
Why are the above true? To find out, change the logarithm to an exponential: $\log_b1=0$logb1=0 becomes $b^0=1$b0=1 and $\log_bb=1$logbb=1 becomes $b^b=1$bb=1
Now, lets review the inverse property:
$\log_bb^x=x$logbbx=x
$b^{\log_bx}=x$blogbx=x$x>0$x>0
Why is this true? Convert each into its opposite: $\log_bb^x=x$logbbx=x becomes $b^x=b^x$bx=bx and $b^{\log_bx}=x$blogbx=x becomes $\log_bx=\log_bx$logbx=logbx
One-to-one
$\log_bM=\log_bN$logbM=logbN if and only if $M=N$M=N
This property is used to solve equations.
Remember, all properties associated with logarithms can be applied to natural logarithms.
#### Worked example
##### Question 1
What is the value of $x$x for the equation $\log_33x=\log_3\left(2x+5\right)$log33x=log3(2x+5)?
Think: Does the logarithms have the same base? Remember, we can not work with logarithms with unlike bases. They do have the same base. So, that means, just as with exponential equations with the same bases, we can set the arguments equal to each other.
Do:
$\log_33x$log33x $=$= $\log_3\left(2x+5\right)$log3(2x+5) (Given) $3x$3x $=$= $2x+5$2x+5 (Use one-to-one property to set arguments equal) $x$x $=$= $5$5 (Subtract both sides by $2x$2x)
### Product rule for logarithms
Recall the product rule for exponents:
$a^na^m=a^{n+m}$anam=an+m
The product rule for logarithms is similar:
$\log_bMN=\log_bM+\log_bN$logbMN=logbM+logbN
Why is this true? Lets use the rules we already know to prove this:
$\log_bMN$logbMN $=$= $\log_bM+\log_bN$logbM+logbN (Given) $b^{\log_bM+\log_bN}$blogbM+logbN $=$= $MN$MN (Rewrite as an exponential) $b^{\log_bM}b^{\log_bN}$blogbMblogbN $=$= $MN$MN (Use laws of exponent: product rule) $MN$MN $=$= $MN$MN (Use inverse property for logarithms)
The product rule for logarithms can be applied repeatedly. The expression $\log_btuv$logbtuv can be rewritten as $\log_bt+\log_bu+\log_bv$logbt+logbu+logbv.
#### Worked example
##### question 2
Rewrite $\log_56x$log56x as two logarithms.
Think: Since there is a product in the logarithm, we can use the product rule for logarithms.
Do: So using the product rule for logarithms, we can rewrite $\log_56x$log56x in the form:
$\log_56+\log_5x$log56+log5x
### Quotient rule for logarithms
Recall the quotient rule for exponents:
The quotient rule for logarithms is similar:
$\log_b\left(\frac{M}{N}\right)=\log_bM-\log_bN$logb(MN)=logbMlogbN
Why is this true? Lets take a look using properties we already know:
$\log_b\left(\frac{M}{N}\right)$logb(MN) $=$= $\log_bM-\log_bN$logbM−logbN (Given) $b^{\log_bM-\log_bN}$blogbM−logbN $=$= $\frac{M}{N}$MN (Rewrite as an exponential) $\frac{b^{\log_bM}}{b^{\log_bN}}$blogbMblogbN $=$= $\frac{M}{N}$MN (Use laws of exponent: quotient rule) $\frac{M}{N}$MN $=$= $\frac{M}{N}$MN (Use inverse property for logarithms)
#### Worked example
##### question 3
Rewrite the logarithmic expression $\log_310-\log_32$log310log32 as one logarithmic expression.
Think: Since the two logarithms have the same base and they are subtracting, we can use the quotient rule for logarithms.
Do: To use the quotient rule for logarithms, we can divide the arguments:
$\log_310-\log_32$log310−log32 (Given) $\log_3\left(\frac{10}{2}\right)$log3(102) (Use quotient rule for logarithms) $\log_35$log35 (Simplify the argument)
#### Practice questions
##### question 4
Simplify each of the following expressions without using a calculator. Leave answers in exact form.
1. $\log_{10}11+\log_{10}2+\log_{10}9$log1011+log102+log109
2. $\log_{10}12-\left(\log_{10}2+\log_{10}3\right)$log1012(log102+log103)
##### question 5
Simplify each of the following expressions without using a calculator. Leave answers in exact form.
1. $\log_{10}18-\log_{10}3$log1018log103
2. $\log_{10}7-\log_{10}28$log107log1028
##### question 6
Express $\log\left(\frac{pq}{r}\right)$log(pqr) as the sum and difference of log terms.
### Power rule for logarithms
We have already seen how to simplify logarithms using the product and quotient properties. Through the definition of logarithms we know that $x=a^m$x=am and $m=\log_ax$m=logax are equivalent. We are able to use this definition to discover some more helpful properties of logarithms such as the power property.
#### Exploration
Let's simplify $\log_a\left(x^2\right)$loga(x2) using the logarithmic properties that we already know.
$\log_a\left(x^2\right)$loga(x2) (Given) $\log_a\left(x\times x\right)$loga(x×x) (Rewrite $x^2$x2as a product, $x\times x$x×x) $\log_ax+\log_ax$logax+logax (Use the product rule for logarithms, $\log_a\left(xy\right)=\log_ax+\log_ay$loga(xy)=logax+logay $2\log_ax$2logax (Collect logarithms with the same base and variables.)
We can also simplify logarithms with powers using the power rule for logarithms, this property can be used for any values of the power $n$n.
Power property of logarithms
$\log_a\left(x^n\right)=n\log_ax$loga(xn)=nlogax
#### Worked example
##### question 7
Rewrite $\log_2\left(x^b\right)$log2(xb) using properties of logarithms. Write your answer without any powers.
Think: The subject of the logarithm has a power, this means we can use the power rule for logarithms,$\log_a\left(x^n\right)=n\log_ax$loga(xn)=nlogax.
What do the values of $a$a and $n$n represent?
Do: In this case $a=2$a=2 and $n=b$n=b. So we can bring the power down to the front, and then multiply it with the logarithm.
$\log_2\left(x^b\right)$log2(xb) $=$= $b\log_2x$blog2x
#### Practice questions
##### QUESTION 8
Use the properties of logarithms to rewrite the expression $\log_4\left(x^7\right)$log4(x7).
##### Question 9
Use the properties of logarithms to rewrite the expression $\log\left(\left(x+6\right)^5\right)$log((x+6)5).
##### QUESTION 10
Use the properties of logarithms to rewrite $\log\left(\left(3x\right)^5\right)$log((3x)5) as the sum of two logarithms.
### Condensing and distributing logarithmic expressions
When viewed together, the product rule, quotient rule and power rule are often called "laws of logs." We can apply more than one rule to simplify an expression.
#### Worked examples
##### question 11
Simplify the expression $\log_3\left(100x^3\right)-\log_3\left(4x\right)$log3(100x3)log3(4x), writing your answer as a single logarithm.
Think: Each logarithm in the expression has the same base, so we can express the difference as a single logarithm using the quotient rule.
Do: To use the quotient rule, we divide the two arguments as follows:
$\log_3\left(100x^3\right)-\log_3\left(4x\right)$log3(100x3)−log3(4x) (Given) $\log_3\left(\frac{100x^3}{4x}\right)$log3(100x34x) (Use the quotient rule) $\log_3\left(25x^2\right)$log3(25x2) (Simplifying the argument) $\log_3\left(\left(5x\right)^2\right)$log3((5x)2) (Rewriting the argument as a power) $2\log_3\left(5x\right)$2log3(5x) (Using the power rule)
##### question 12
Distribute the expression: $\ln\frac{x^2y^3}{z^4}$lnx2y3z4
Think: List the operations within the argument: exponents, multiplication, division. We are going to start distributing with division because of the sensitivity to order.
Do:
$\ln\frac{x^2y^3}{z^4}$lnx2y3z4 (Given) $\ln x^2y^3-\ln z^4$lnx2y3−lnz4 (Use the quotient rule) $\ln x^2+\ln y^3-\ln z^4$lnx2+lny3−lnz4 (Use the product rule) $2\ln x+3\ln y-4\ln z$2lnx+3lny−4lnz (Use power rule)
#### Practice questions
##### Question 13
Express $3\ln\left(x^5\right)-4\ln\left(x^2\right)$3ln(x5)4ln(x2) as a single log expression.
##### question 14
Express $5\log x+3\log y$5logx+3logy as a single logarithm.
## Changing the base
We often encounter occasions where we need to take a logarithm given in one base and express it as a logarithm in another base. A change of base formula has been developed to do just that.
Change of Base Rule!
$\log_ab=\frac{\log_cb}{\log_ca}=\frac{1}{\log_ba}$logab=logcblogca=1logba
#### Worked examples
##### question 15
Change the base of the logarithm to $2$2: $\log_48$log48 and evaluate.
Think: Use the change of base rule.
Do:
$\log_48$log48 (Given) $\frac{\log_28}{\log_24}$log28log24 (Use the change of base rule) $\frac{\log_22^3}{\log_22^2}$log223log222 (Rewrite the argument as exponents) $\frac{3}{2}$32 (Use the inverse property)
##### question 16
Evaluate $\log_9564$log9564 using by changing the base to $10$10 . Round to 3 decimal points.
Think: Review the change of base rule. Can we rewrite the current base or argument as exponents with base $10$10?
Do:
$\log_9564$log9564 (Given) $\frac{\log564}{\log9}$log564log9 (Use change of base rule) $2.883$2.883 (Cannot write the arguments as exponents with base $10$10, use calculator to find $\log564$log564 and $\log9$log9 then divide)
#### Practice questions
##### QUESTION 17
Rewrite $\log_416$log416 in terms of base $10$10 logarithms.
##### QUESTION 18
Rewrite $\log_320$log320 in terms of base $4$4 logarithms.
##### QUESTION 19
Rewrite $\log_3\sqrt{5}$log35 in terms of base $10$10 logarithms. |
The 4th term of a G.P. is square of its second term, and the first term is –3. Determine its 7th term.
Asked by Abhisek | 1 year ago | 84
##### Solution :-
Let’s consider a to be the first term and r to be the common ratio of the G.P.
Given, a = –3
And we know that,
an = arn–1
So, a= ar3 = (–3) r3
a2 = a r1 = (–3) r
Then from the question, we have
(–3) r3 = [(–3) r]2
⇒ –3r3 = 9 r2
⇒ r = –3
a7 = a r 7–1 = a r6 = (–3) (–3)6 = – (3)7 = –2187
Therefore, the seventh term of the G.P. is –2187.
Answered by Pragya Singh | 1 year ago
### Related Questions
#### Construct a quadratic in x such that A.M. of its roots is A and G.M. is G.
Construct a quadratic in x such that A.M. of its roots is A and G.M. is G.
#### Find the two numbers whose A.M. is 25 and GM is 20.
Find the two numbers whose A.M. is 25 and GM is 20.
#### If a is the G.M. of 2 and 1/4 find a.
If a is the G.M. of 2 and $$\dfrac{1}{4}$$ find a.
#### Find the geometric means of the following pairs of numbers
Find the geometric means of the following pairs of numbers:
(i) 2 and 8
(ii) a3b and ab3
(iii) –8 and –2
Insert 5 geometric means between $$\dfrac{32}{9}$$ and $$\dfrac{81}{2}$$. |
How do you find the vertical, horizontal or slant asymptotes for f(x) = (-x^2 + 4x)/(x+2)?
Jan 6, 2017
The vertical asymptote is $x = - 2$
The slant asymptote is $y = - x + 6$
No horizontal asymptote
Explanation:
The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- 2\right\}$
As we cannot divide by $0$, $x \ne - 2$
The vertical asymptote is $x = - 2$
The degree of the numerator is $>$ than the degree of the denominator, so there is a slant asymptote.
To find the slant, we start by doing a long division
$\textcolor{w h i t e}{a a a a}$$- {x}^{2} + 4 x$$\textcolor{w h i t e}{a a a a}$∣$x + 2$
$\textcolor{w h i t e}{a a a a}$$- {x}^{2} - 2 x$$\textcolor{w h i t e}{a a a a}$∣$- x + 6$
$\textcolor{w h i t e}{a a a a a a a}$$0 + 6 x$
$\textcolor{w h i t e}{a a a a a a a a a}$$+ 6 x + 12$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a}$$- 12$
So,
$f \left(x\right) = \frac{- 4 {x}^{2} + 4 x}{x + 2} = - x + 6 - \frac{12}{x + 2}$
${\lim}_{x \to - \infty} f \left(x\right) + \left(x - 6\right) = {\lim}_{x \to - \infty} - \frac{12}{x + 2} = {0}^{+}$
${\lim}_{x \to + \infty} f \left(x\right) + \left(x - 6\right) = {\lim}_{x \to + \infty} - \frac{12}{x + 2} = {0}^{-}$
The slant asymptote is $y = - x + 6$
No horizontal asymptote
graph{(y-(-x^2+4x)/(x+2))(y+x-6)(y-25x-50)=0 [-35, 38.04, -14.36, 22.2]} |
#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 65
$\frac{x^{2}}{2}+x+\frac{1}{2} \log |x-1|-\frac{1}{4} \log \left|x^{2}+1\right|-\frac{1}{2} \tan ^{-1} x+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x^{4}}{(x-1)\left(x^{2}+1\right)} d x$
Explanation:
Let
\begin{aligned} &I=\int \frac{x^{4}}{(x-1)\left(x^{2}+1\right)} d x \\ &I=\int \frac{x^{4}}{x^{3}-x^{2}+x-1} d x \\ &I=\frac{x\left(x^{3}-x^{2}+x-1\right)+1\left(x^{3}-x^{2}+x-1\right)+1}{\left(x^{3}-x^{2}+x-1\right)} \\ &I=x+1+\frac{1}{(x-1)\left(x^{2}+1\right)} \end{aligned}
Let,
\begin{aligned} &\frac{1}{(x-1)\left(x^{2}+1\right)}=\frac{A}{x-1}+\frac{B x+C}{x^{2}+1} \\ &1=A\left(x^{2}+1\right)+(B x+C)(x-1) \end{aligned}
Put $x=1$
\begin{aligned} &1=2 A \\ &A=\frac{1}{2} \end{aligned}
Put $x= 0$
\begin{aligned} &1=A-C \\ &C=A-1=-\frac{1}{2} \\ &C=\frac{-1}{2} \end{aligned}
Put $x= -1$
\begin{aligned} &1=2 A+2 B-2 C \\ &1=2(A-C)+2 B \\ &1=2+2 B \\ &2 B=-1 \\ &B=\frac{-1}{2} \end{aligned}
\begin{aligned} &\int \frac{x^{2}}{(x-1)\left(x^{2}+1\right)} d x=\int x d x+\int 1 d x+\frac{1}{2} \int \frac{1}{x-1} d x-\frac{1}{2} \int \frac{x+1}{x^{2}+1} d x \\ &=\frac{x^{2}}{2}+x+\frac{1}{2} \log |x-1|-\frac{1}{4} \log \left|x^{2}+1\right|-\frac{1}{2} \tan ^{-1} x+C \end{aligned} |
# Pre-teaching Slope with Scratch
I gave my 6th graders a challenge in Scratch today and loved the potential to build background on slope as they go into algebra classes later.
This is a starter program in which a cat uses the pen tools to draw a set of stairs to reach a princess.
https://scratch.mit.edu/projects/58903244/#editor
Using the pen tools so the cat can reach his friend princess using stairs.
The instructions are:
The cat asks you for the height and width of each step. It should draw an entire set of steps using a loop. Test to see if the cat reaches the princess. Can you guess a height and width for the steps that will reach the princess exactly?
I loved watching the students interact with the project. There were several different solutions that came out, and several different misconceptions. It was fairly common for students to plug random “move” blocks into a loop and test it without really analyzing why they were doing what they did. Getting the students to think through the sequence that needed to be repeated was challenging. Move up. Up how much? then turn. Move right. right how much? turn. Repeat. Some students used a move-turn-move-turn sequence and others figured out it took fewer blocks if you used change x – change y. Some tried to work with “glide” blocks, but since they had trouble thinking through the math of the new coordinates, nobody completed it.
Once they created the steps, they tried to figure out how to get the cat on a collision course with the princess. It was fascinating. It was very common for students to try height = 10, width = 10. The cat would travel at a 45 degree angle and end up above the princess. So they would try height = 20, width = 20. When that didn’t work, height = 50, width = 50. At this point, many would start to try entering different values for the height and width. If the height is greater than the width, the cat overshoots the princess even more. But if the height is anywhere from two-fifths to three-fifths of the width, the cat would collide with the princess. A ratio of 5/12 works. 20/30 just barely misses. 10/20 works. 6/10 works but just barely. 8/10 does not work. Through trial and error, students figured out a height and width for the steps that reaches the princess.
I thought about how this experience might be useful prior to the students’ learning about slope. I had forgotten that Common Core doesn’t emphasize the rise/run ratio as a method for teaching slope, but focuses on unit rates:
CCSS.MATH.CONTENT.8.F.B.4
Construct a function to model a linear relationship between two quantities. Determine the rate of change and initial value of the function from a description of a relationship or from two (x, y) values, including reading these from a table or from a graph. Interpret the rate of change and initial value of a linear function in terms of the situation it models, and in terms of its graph or a table of values.
Here you see the ratio understanding coming out in 8th grade, but the ideas around ratios build all the way through middle school.
CCSS.MATH.CONTENT.8.EE.B.6
Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane; derive the equation y = mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b.
I would love to have the students gather data on what height and width were needed to get the cat to the princess and make some inferences about the relationship between height, width, and coordinates. From there, wouldn’t it be interesting to modify the activity such that the cat and princess start at random locations each time and you need to calculate the height and width of the stairs?
An additional layer of challenge would be to limit the students to only using the “move to x: y:” and “glide to x: y:” blocks – so they have do to the math on the x and y coordinates for the stairs. It’s interesting how working with addition and subtraction as motion in the coordinate plane really expands your knowledge of those operations – we work so often with the idea of “put together” and “take away” that we forget they can represent vectors of motion.
I'm a former software engineer who has taught middle school math and computer science for the past 6 years. I believe every kid has the right to be a thinker. I started this blog to save resources for integrating programming in the Common Core math classroom. I also use it to save my lessons and reflections from teaching budding computer scientists! Coding has transformed how I teach and think. You'll love what it does for you. You should try it.
### 2 responses to “Pre-teaching Slope with Scratch”
1. zamanskym says :
Nice.
There’s got to be a cool netlogo assignment in here as well.
2. Joe says :
I LOVE the idea of introducing slope like this. I think it needs to go farther though. In your example, slope is a ratio between two distances (said more plainly, a number). As a science teacher, I struggle with students understanding slope as a rate of change. I’d love to see a version of this that requires a relationship between two different units, like money and time. There could be a target amount on a certain day, like \$500 on day 10. Then students could do the same thing with a \$ vs day graph. Students then find a slope in \$/day. I think that may reinforce the rate of change idea. To extend that example, you could have a target amount, like \$500 on day 15, and a starting balance, like \$125 on day 0. I think this might help prevent the mistake that I see in physics so often (taking a single position at a certain time and solving speed by dividing the two).
tl;dr: Continue the programming challenge to include more meaningful variables and extend it to include a y-intercept. |
# Show that there are no two positive integers $x$ and $y$ such that $x^3=2^y+15$.
Show that there are no two positive integers $$x$$ and $$y$$ such that $$x^3=2^y+15$$.
Attempt
For the sake of contradiction, suppose that there are two positive integers $$x$$ and $$y$$ satisfying $$x^3 = 2^y + 15$$.
Consider the equation modulo $$4$$. The cubes of integers modulo $$4$$ can only yield remainders of $$0$$, $$1$$, or $$3$$. This can be verified by calculating the cubes of the numbers $$0$$, $$1$$, $$2$$, and $$3$$ modulo $$4$$.
Now, let's analyze the possible remainders of powers of $$2$$ modulo $$4$$. We have $$2^0 \equiv 1 \pmod{4}$$, $$2^1 \equiv 2 \pmod{4}$$, $$2^2 \equiv 0 \pmod{4}$$, and $$2^3 \equiv 0 \pmod{4}$$. As we can see, for $$y \geq 2$$, $$2^y$$ is divisible by $$4$$.
Using this information, let's consider the equation $$x^3 = 2^y + 15$$ modulo $$4$$. We have two cases:
Case 1: $$y = 1$$ If $$y = 1$$, then $$2^y = 2$$, and the equation becomes $$x^3 = 2 + 15$$. This simplifies to $$x^3 = 17$$. However, no positive integer cubed equals $$17$$. This contradicts the equation, so this case is not possible.
Case 2: $$y \geq 2$$ If $$y \geq 2$$, then $$2^y$$ is divisible by $$4$$. Adding $$15$$ to $$2^y$$ yields a number that leaves a remainder of $$3$$ when divided by $$4$$. However, as mentioned earlier, the cubes of integers modulo $$4$$ can only yield remainders of $$0$$, $$1$$, or $$3$$. Therefore, there are no positive integers $$x$$ and $$y$$ that satisfy the equation $$x^3 = 2^y + 15$$ in this case.
Since we have considered all possible cases and found contradictions in each case, we can conclude that there are no positive integers $$x$$ and $$y$$ that satisfy the equation $$x^3 = 2^y + 15$$. Thus, the assumption that such integers exist must be false.
Hence, we have proven by contradiction that there are no two positive integers $$x$$ and $$y$$ satisfying the equation $$x^3 = 2^y + 15$$. Q.E.D.
$${}$$
$${}$$
Second Attempt:
Suppose, for the sake of contradiction, that there are two positive integers $$x$$ and $$y$$ such that $$x^3=2^y+15$$.
Notice that for any positive integer $$x$$, we have $$x^3 \equiv 0,1,6 \pmod 7$$.
On the other hand, for any positive integer $$y$$, we have $$2^y \equiv 1,2,4 \pmod 7, \tag{1}$$ i.e.,$$2^y+15 \equiv 2,3,5 \pmod 7$$.
Comparing the congruence relations, we see that there is no overlap between the possible values of the left side ($$0,1,6$$) and the possible values of the right side ($$2,3,5$$) modulo $$7$$.
Hence, our assumption is false, which means that there are no two positive integers $$x$$ and $$y$$ such that $$x^3=2^y+15$$. Q.E.D.
$${}$$
• Where do have doubts? You already had this comment in your last question: For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. May 23, 2023 at 8:49
• In case 2) you say that $15+2^y$ has remainder $3$, then say that $3$ is also a possible remainder of the cube. May 23, 2023 at 8:50
• Shouldn't $2^y+15\equiv2,3,5\mod7$ instead of $0,2,5$? But that arrives at the conclusion even faster (since $x^3\equiv0,1,6\mod7$ doesn't match at all), good job 🙂 May 23, 2023 at 9:49
• @TheMather-orratherAMather Ah, my bad. Thanks for pointing out my mistake. So, the proof was correct? May 23, 2023 at 9:58
• @math404 yes, it's correct👌 May 23, 2023 at 10:04
The second attempt works when you render the modulo $$7$$ results correctly:
$$x^3\in\{0,1,6\}$$
$$2^y\in\{1,2,4\}\implies2^y+15\in\{2,3,5\}$$.
$$\rightarrow\leftarrow$$
Case 2 is handled incorrectly. You basically say:
1. If $$y\geq 2$$, then $$2^y+15$$ has remainder $$3$$ when divided by $$4$$.
2. Any cube must have remainder $$0,1$$ or $$3$$ when divided by $$4$$.
3. Therefore, $$2^y+15$$ is not a cube.
But the mistake there is, quite simply, that the conclusion (3) does not follow from the premises (1 and 2).
• So, how to handle it? May 23, 2023 at 8:57
• what about my second attempt? May 23, 2023 at 9:45 |
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Visualizing Solid Shapes
Any plane or any shape has two measurements like length and width, that is why it is called a two-dimensional(2D) object. Circle, square, triangles, rectangles, trapeziums, etc. are 2-D shapes. If an object has length, width and breadth then it is a three-dimensional object(3D). cube, pyramids, spheres, cylinders, cuboids are 3-D shapes. Any kind of solid shape occupies some space. A solid shape or figure is bounded by one or more surfaces. If any two faces of 3-D shapes meet together, we get a line segment which is called edge when more than two faces of the solid meet at one point then that point is called the vertex of solid. A 3-D solid have different views from different positions. A two-dimensional shape and three-dimensional shape can be named 2D and 3D objects.
Now, you can see some 2D and 3D objects are given below:
2-D Shapes
3D -Shapes
Faces, Edges, and Vertices
Face
A face refers to any single flat surface of a solid object. Solid shapes can have more than one face. The polygonal regions which a solid is made of are called faces.
Edges
An edge is a line segment on the boundary joining one vertex (corner point) to another. They serve as the junction of two faces. The faces meet at edges which are lines.
Vertices
A point where two or more lines meet is called a vertex. It is a corner. The point of intersection of edges denotes the vertices. These edges meet at vertices which are points
Now, if we talk about three-dimensional shapes then they have different numbers of faces, vertices, and edges. All that flat surface of shape is called a face. This flat shape is two-dimensional. The line segment where the faces of three-dimensional shapes meet each other is called the edge of the figure. The points or the corners where edges meet each other are called vertices.
Now, if we classified the number of faces, vertex and edge then this cuboid has 6 faces, 12 edges, and 8 vertices.
View of 3D Shapes
Any three-dimensional figure or shape has a top view, side view, and front view.
Top View: Shape of the object when you see the object from the top or from directly above called as Top view of an object.
Side View: Shape of the object when you see the object from one side as mentioned in the below figure.
Front View: Shape of the object when you see the object from the front direction as mentioned in the below figure.
Now look at the cube, we will look at the top view, side view, and front view
Front View:
All we see a front face which is a square.
Top View:
All we see a top view which is also a square.
Side View:
This is the side view (left and right) of the cube which also a square.
Note: A Cube will always look like a square whether its front view, side view or top view.
Cylinder
A cylinder is a three-dimensional solid that contains two parallel bases connected by a curved surface. The bases are usually circular in shape. The perpendicular distance between the bases is denoted as the height “h” of the cylinder and “r” is the radius of the cylinder.
Top view: When we see the cylinder from the top then it looks like a circle.
Circle
Front View: When we see the cylinder from the front view then its looks like a rectangle.
Front View
Side View: When we see the cylinder from the side view then its looks like a rectangle.
Side View
Square Pyramid
A pyramid is a 3-dimensional geometric shape formed by connecting all the corners of a polygon to a central apex.
There are many types of pyramids. Most often, they are named after the type of base they have. Following is a square pyramid because of its base as a square.
Side view of pyramid will look like a triangular shape for left and right side.
Bottom of the pyramid has square shape.
Faces = 5
Edges = 8
Vertices = 5
Triangular Pyramid
Side view of the pyramid will look like a triangular shape for the left and right sides.
Bottom of the pyramid has a triangle shape.
Faces = 4
Edges = 6
Vertices = 4
Cone
Cone
Side view for cone, it will look like a triangle.
From the top, it will look like a circle.
Faces = 2
Edges = 2
Vertices = 1
Sample Questions
Question 1: How many vertices are there in a sphere?
Sphere has no vertices in there because it has round shape.
Sphere
Question 2: Is a cone polyhedron? Give explanation. |
# Lesson 15
Multiply More Fractions
## Warm-up: Number Talk: Multiply Mixed Numbers (10 minutes)
### Narrative
The purpose of this Number Talk is for students to demonstrate strategies and understandings they have for using the properties of operations when multiplying whole numbers and mixed numbers. These understandings help students develop fluency and will be helpful later in this lesson when students will flexibly multiply.
### Launch
• Display one problem.
• “Give me a signal when you have an answer and can explain how you got it.”
• 1 minute: quiet think time
### Activity
• Keep problems and work displayed.
• Repeat with each problem.
### Student Facing
Find the value of each expression mentally.
• $$6 \times \frac {3}{8}$$
• $$6 \times 2 \frac {3}{8}$$
• $$7 \times \frac {9}{10}$$
• $$7 \times 3 \frac {9}{10}$$
### Activity Synthesis
• Display the last problem.
• “If someone found $$7 \times 4$$ first, what might they do next?” (Subtract $$7 \times \frac{1}{10}$$ since $$3 \frac{9}{10}$$ is $$\frac{1}{10}$$ less than 4.)
## Activity 1: Multiply Your Way (20 minutes)
### Narrative
The purpose of this activity is for students to consider situations and the operations involved in order to select reasonable numbers for each situation. Students are given a set of numbers which must each be used once in the statements. Students may find that their initial thinking does not work with the given constraints and may need to revise their work (MP1). They also need to think carefully about the units involved and consider whether a number is representing a linear unit or a square unit. Students may solve these problems using several different strategies.
• Groups of 2
### Activity
• 3–5 minutes: independent work time
• 8–10 minutes: partner discussion
• Monitor for students who:
• draw an area diagram.
• write multiplication equations.
• revise their thinking and can explain why their original solution didn’t work, but their revised solution does work.
### Student Facing
Write numbers from the list in the blank spaces so the situations make sense. Each number will be used only one time. Be prepared to explain your thinking.
• 4
• 5
• $$5 \frac {1}{2}$$
• 3
• $$5\frac {3}{4}$$
• 2
1. The area of the rug is $$16 \frac {1}{2}$$ square feet. The length of the rug is ___________ feet.
The width of the rug is ___________ feet.
2. The puzzle is $$2 \frac {1}{2}$$ feet wide. It is _______ feet long. It has an area of _________ square feet.
3. The area of the whiteboard is 23 square feet. The length of the whiteboard is
_________ feet. The width of the whiteboard is ________ feet.
### Student Response
If students need more of an invitation to enter the task, display the first problem with no numbers. Ask students to name some numbers that would make sense in this situation and explain why.
### Activity Synthesis
• Ask previously selected students to share their solutions.
• “Which numbers did you think made sense for the length of the rug in feet?” (I thought any of the numbers 5, $$5\frac{1}{2}$$, or $$5\frac{3}{4}$$ made sense.)
• “How did you decide which number to use for the length of the rug?” (The product had to be $$16\frac{1}{2}$$. So I tried a width of 2 feet and that was too small. Then I tried a width of 3 feet and that worked with $$5\frac{1}{2}$$ feet for the length.)
## Activity 2: Equivalent Expressions (15 minutes)
### Narrative
The purpose of this activity is for students to match different diagrams and expressions representing the same product. In the previous several lessons, students have studied different ways to find products of a whole number and a fraction using arithmetic properties such as the distributive property. Earlier in the unit they learned about the connection between fractions and division. They combine these skills as the expressions and diagrams they work with incorporate both the distributive property and the interpretation of a fraction as division (MP7). Students may match diagrams to expressions differently than the ways that are listed in the sample responses. Encourage students to match the expressions and diagrams in a way that makes sense to them, as long as they can accurately explain how the expression is represented in the diagram they chose.
MLR7 Compare and Connect. Invite students to prepare a visual display that shows their thinking about their favorite diagram and expression. Encourage students to include details that will help others understand what they see, such as using different colors, arrows, labels, or notes. If time allows, invite students time to investigate each others’ work.
Engagement: Develop Effort and Persistence. Chunk this task into more manageable parts. Give students a subset of options to start with and introduce the remaining numbers once students have completed their initial set of matches.
Supports accessibility for: Organization, Social-Emotional Functioning
### Launch
• Groups of 2
• “You are going to match expressions and diagrams that show different ways to find the value of the product $$4 \times 5 \frac{2}{3}$$.”
### Activity
• 5–7 minutes: independent work time
• 2–3 minutes: partner discussion
### Student Facing
Each diagram represents a way to calculate $$4 \times 5 \frac{2}{3}$$. Each expression is equivalent to $$4 \times 5 \frac{2}{3}$$. Match the diagrams and expressions. Show or explain your reasoning.
1. $$(4 \times 5) + \left(4 \times \frac{2}{3}\right)$$
2. $$(4 \times 6) - \left(4 \times \frac{1}{3}\right)$$
$$\phantom{\hspace{2.5cm} \\ \hspace{2.5cm}}$$
3. $$4 \times \frac{17}{3}$$
$$\phantom{\hspace{2.5cm} \\ \hspace{2.5cm} \\ \hspace{2.5cm}}$$
4. $$(4 \times 17) \div 3$$
Choose your favorite diagram and expression to find the value of $$4 \times 5 \frac{2}{3}$$. Explain why it is your favorite.
### Activity Synthesis
• Display the diagrams B and D.
• “How are the diagrams the same? How are they different?” (They both have the same shaded area and the same divisions inside. The top one shows the length of the shaded rectangle. The bottom one does not though it has enough information to find the shaded area.)
• Invite students to share their favorite ways to find the product.
## Lesson Synthesis
### Lesson Synthesis
“Today we used what we have learned to find the value of expressions involving multiplication of whole numbers and fractions greater than 1 written as mixed numbers.”
Display: $$7 \frac {3}{5} \times 6$$
“Tell me everything you know about this expression.” (It is equal to a number between 42 and 48. It is equal to $$\frac {38}{5} \times 6$$ and to $$\left(7 + \frac{3}{5}\right) \times 6$$. We can draw an area diagram to represent $$7 \frac {3}{5} \times 6$$.)
Record student responses for all to see.
If not mentioned by students, record an area diagram and equivalent expressions. |
# Rotation of Axes (Change of Direction)
### Rotation of Axes (Change of Direction)
1) When the axes rotated through an angle 60, then new co-ordinates of three points are the following (3, 4).
Solution: Given that,
New co-ordinates are (3, 4)
X = 3, Y = 4
x = X cos θ – Y sin θ
= 3 cos60 – 4 sin60
= 3 (½) – 4 $$\frac{\sqrt{3}}{2}$$,
= $$\frac{3-4\sqrt{3}}{2}$$.
y = X sin θ + Y cos θ
= 3 sin60 + 4 cos60
= 3 $$\frac{\sqrt{3}}{2}$$ + 4 (½)
= $$\frac{4+\sqrt{3}}{2}$$.
Co-ordinate of P are $$\left( \frac{3-4\sqrt{3}}{2},\frac{4+\sqrt{3}}{2} \right)$$.
2) Find the angle through which the axes are to be rotated so as to remove the xy term in the equation x² + 4xy + y² – 2x + 2y – 6 = 0.
Solution: Comparing the equation
x² + 4xy + y² – 2x + 2y – 6 = 0 with ax² + 2hxy + by² + 2gx + 2fy + c = 0
a = 1, h = 2, b = 1, g = -1, f = 1, c = -6
let θ be the angle of rotation of axes, then θ = ½ tan⁻¹$$\left( \frac{2h}{a-b} \right)$$,
= ½ tan⁻¹ (4/ 1 – 1) = ½ tan⁻¹ (4/0)
= ½ tan⁻¹(∞) = ½ x π/2
θ = π/4
3) When the axes are rotated through an angle 45, the transformed equation of a curve is 17x² – 16xy + 17y² = 255. Find the original equation of the curve?
Solution: Angle of rotation = θ = 45
X = x cosθ + y sinθ
= x cos45 + y sin45 = (x + y)/2
Y = -x sinθ + y cosθ
= -xsin45 + ycos45 = (-x + y)/2
The original equation is 17x² – 16xy + 17y² = 255
⇒ $${{17\left( \frac{x+y}{2} \right)}^{2}}-16\left( \frac{x+y}{2} \right)\left( \frac{-x+y}{2} \right)+17{{\left( \frac{-x+y}{2} \right)}^{2}}=225$$,
⇒ $$17\left( \frac{{{x}^{2}}+{{y}^{2}}+2xy}{2} \right)-16\left( \frac{{{x}^{2}}-{{y}^{2}}}{2} \right)+17\left( \frac{{{x}^{2}}+{{y}^{2}}+2xy}{2} \right)=225$$,
⇒ 17 (x² + y²) – 8 (x² – y²) = 225
⇒ 19 x² + 25 y² = 225. |
Exploration of Some Polar Equations
by Asli Ersoz
In this write-up I investigate the polar equation .
First, I set a = b = 1 and vary k through integer values.
For k = 0, the graph looks like
This makes sense because for k = 0, the equation becomes r = 1 + cos (0) = 2 and whatever the angle is r will be 2. So, the graph will be the set of points which are 2 units away from the origin.
And for k = 1, it looks like
Before going on with other graphs you can click here and onserve how the graph changes as k varies.
Here are the graphs for some other values of k:
So, one thing that we can obseve and conjecture from these graphs is that k determines the number of leaves or the petals.
We observe another thing if we look at all the graphs in one picture:
The graph for k = 0, in other words the circle, seems to be the boundary for all others. Why?
This can be explained with a characteristic of the cosine function. Cosine of an angle will always be between -1 and 1. Since, it can't exceed 1, in our polar equation r won't exceed 2. So, none of the graphs will go beyond the circle.
Now, let's see what happens when k takes a negative value:
These look exactly the same as the previous graphs. Why? Again because of the cosine function. Cosine of an angle is equal to the cosine of the negative of the same angle.
Now, let's compare the graph of
with the graph of
.
For k = 0, the obvious is:
For k = 1:
Here, I wondered why the second one became a circle and I decided to play around withand see what it looks like as an equation in x and y.
The correspondance between polar corrdinates and the rectangular coordinates is like the following:
So,
Applying some trigonometric laws:
From here, we can see that
And, this is exactly the equation for the green circle!
Now, let's go back and compare our functions for other values of k:
So, a conjecture that we might have at this point is when k is odd graphs of the two functions have equal number of leaves and when k is even the graph of has twice as many number of leaves as the graph of .
Some other questions that could be explored:
Why do these functions behave the way they do? Why does k change the number of leaves?
What happens when a and b are also varied?
What happens when k takes non-integer values?
What happens if cosine function is replced with the sine function? |
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# Learner Guide
Maths
Number
Page: 1 | 2 < >
Equivalent Fractions
Fractions represent a portion of the whole. Different fractions which represent
an equal amount of the whole are called equivalent fractions.
In order to find different equivalent fractions, we can multiply or divide both
the numerator and denominator of the fractions we already have.
For example, if we wanted to find an equivalent fraction of 3/4, we could do
this multiplying both numbers by two, which would give us 6/8.
Example 1. Take a look at this cake. We can cut it into different numbers of slices and still represent one half. In the first circle, we can show a half by taking one piece out of a possible two, so this is 1/2. In the second circle, one half is shown by highlighting 2 pieces out of a possible four, so this is 2/4. The third circle is 3/6 and the fourth circle is 4/8. As all of these fractions represent the same amount of space, we would say that they are equivalent fractions. We can see that the numerator and denominator of the first fraction 1/2 have been multiplied by 2, 3 and 4 to get the other fractions.
Equivalent fractions can come in very useful when we need to compare
fractions that have different denominators. If we can use equivalent fractions
to make both fractions have the same denominator, all we have to do is
compare the numerator.
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# 1981 AHSME Problems/Problem 24
## Problem
If $\theta$ is a constant such that $0 < \theta < \pi$ and $x + \dfrac{1}{x} = 2\cos{\theta}$, then for each positive integer $n$, $x^n + \dfrac{1}{x^n}$ equals
$\textbf{(A)}\ 2\cos\theta\qquad \textbf{(B)}\ 2^n\cos\theta\qquad \textbf{(C)}\ 2\cos^n\theta\qquad \textbf{(D)}\ 2\cos n\theta\qquad \textbf{(E)}\ 2^n\cos^n\theta$
## Solution
Multiply both sides by $x$ and rearrange to $x^2-2x\cos(\theta)+1=0$. Using the quadratic equation, we can solve for $x$. After some simplifying:
$$x=\cos(\theta) + \sqrt{\cos^2(\theta)-1}$$ $$x=\cos(\theta) + \sqrt{(-1)(\sin^2(\theta))}$$ $$x=\cos(\theta) + i\sin(\theta)$$
Substituting this expression in to the desired $x^n + \dfrac{1}{x^n}$ gives:
$$(\cos(\theta) + i\sin(\theta))^n + (\cos(\theta) + i\sin(\theta))^{-n}$$
Using DeMoivre's Theorem:
$$=\cos(n\theta) + i\sin(n\theta) + \cos(-n\theta) + i\sin(-n\theta)$$
Because $\cos$ is even and $\sin$ is odd:
$$=\cos(n\theta) + i\sin(n\theta) + \cos(n\theta) - i\sin(n\theta)$$ $=\boxed{\textbf{2\cos(n\theta)}}$ (Error compiling LaTeX. )
Which gives the answer $\boxed{\textbf{D}}$ |
### Lesson 5-4 Point
```Lesson 5-4
Point-Slope Form
Sept. 22, 2014
Daily Learning Target
• I will write and graph linear
equations using point-slope
form.
What is it?
y – y1 = m(x – x1)
m = slope
(x1, y1) = a point the line
goes through.
Here is how it works!!
• Given a point (x1, y1) on a line and the
line’s slope m, you can use the definition
of slope to derive point-slope form.
y2 – y1
• Definition of slope (m) =
x2 – x1
• Let (x, y) be any point on the line.
Substitute for (x, y) for (x2, y2).
y – y1
•
=m
x – x1
How it works cont.
y – y1
•
=m
x – x1
• Multiply both sides by x – x1
y – y1 = m(x – x1 ) point-slope form
m = slope
(x1, y1) = a point the line goes through.
Example 1
• A line passes through (-3, 6) and has slope -5.
What is an equation of the line in point slope
form?
**Remember: y – y1 = m(x – x1 )
y – 6 = -5(x + 3 )
You Try It
2
.
3
• A line passes through (8, -4) and has slope
What is an equation in point-slope for of the
line?
y+4=
2
(x
3
– 8)
Example 2
• What is the graph
of the equation
y–1=
2
(x
3
– 2)
What is the slope?
What is the point
it passes through?
You Try
• What is the graph
of the equation:
a. y + 7 = -
4
(x
5
– 4)
b. y – 1 = -3(x + 2)
c. y – 2 =
4
(x
9
– 3)
Example 3
• What is an equation of the line from the
coordinate plane?
the slope formula.
– Use the given points to find the slope
– Use point-slope form
– Use either point for (x1, y1).
You Try
• Use the same procedure as the example to
write the equation for a graph where the line
goes through points (-3, 4) and (1, 1).
y–1=-
3
(x
4
– 1)
Example 4
• The table shows the altitude of a hot-air
balloon during its linear descent. What
equation in point-slope form. What does it
look like in slope-intercept form?
• PSF: y – 640 = -2.5(x – 10)
• SIF: y = -2.5x + 665
Time, x (s)
Altitude, y (m)
10
640
30
590
70
490
90
440
You Try
• The table shows the number of gallons of
water, y, in a tank after x hours. The
relationship is linear. What is an equation in
point-slope forms that models that data.
What does the slope
Time, x (h)
Water, y (gal)
represent?
2
3320
PSF: y – 3320 = 1250(x – 2);
3
4570
The rate at which water is
5
7070
8
10,820 |
In mathematical terms, the set of necklaces is the set of strings with equivalence under rotation.
A string is a (finite) list of symbols taken from some symbol set. A rotation of a string is any other string that can be formed by splitting the original in two, and appending the first substring to the second (so the two substrings are in the opposite order, but the symbols within each remain in the same order). For example, given the string "judge me by my size do you " (-- Yoda), it can be broken up into "judge me by my size " and "do you ", so "do you judge me by my size " is the same necklace as "judge me by my size do you ". However, "you do judge me by my size " is not the same necklace, because there is no way to rotate the original string to get this string.
When a symbol set is considered to be a set of digits, a string can be interpreted as a number, represented in a base equal to the size of the symbol set (e.g. when the symbol set is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, a string can be interpreted as a decimal integer). The equivalence of rotation then has a numerical interpretation, and as the 7-digit necklace "3579001" is equivalent to the 7-digit necklace "0013579", one can say that in 7-digit necklace values, the number 3,579,001 is equivalent to 13,579). In this manner, one can choose the lowest-valued rotation as a unique representation of a string. This is useful in comparing necklaces to check if they are equivalent; e.g., given two necklaces (say "8000246" and "246800"), find the lowest-valued rotation of each ("0002468" and "0002468") - if the lowest-valued rotations are equal as strings, then the two original values are equal as necklaces. These 'lowest-valued' rotations have the property that, given any substring, the value of that substring is always greater than or equal to the value of the substring of the same length at the beginning of the string. (Otherwise, one could split the string at the beginning of that substring, swap the two parts and get a rotation which has a lower value.)
Necklaces are sometimes referred to as having a number of 'colour's, and a number of 'bead's. The number of colours is the same as the number of symbols in the symbol set. The number of beads is the length of the string. The number of different necklaces with n beads and k colours is given (using Eindhoven notation) by the expression:
(1 / n) * (+ d : d divides n : totient(d) * k(n / d))
A binary necklace is a necklace of two colours, i.e. a necklace in which the symbol set only has two elements (e.g. binary digits). Binary necklaces are commonly used in RFID (Radio Frequency IDentification). An ID tag is programmed with a certain binary string, and it transmits that string over and over again continuously. Thus, if a tag is programmed with the string '10001011', it transmits: '…100010111000101110001011…'. A reader would have no way of identifying where the tag 'starts', so all the reader can do is detect a binary necklace. In this case, the reader would detect the necklace '00010111' (the lowest-valued rotation of the string in the tag).
In order that ID codes in RFID be unique (a fundamental requirement of any ID system), no two tags can be allowed to transmit the same necklace (a stronger requirement than, no two tags can be allowed to transmit the same string). This is usually accomplished by deciding on some synchronisation pattern, and then ensuring that the same pattern appears nowhere else in the transmission. For example, one might choose to have a sync pattern of '000001', and to avoid having that pattern appear anywhere else in the transmission, a '1' could be stuffed after every 4 data bits. This way, the reader can rotate the necklace it receives until the sync pattern is at the beginning of the string, can check that all the stuffing bits, and can extract the data bits. Thus, even though the transmission has no physical starting point, a logical starting point can be determined and data can be communicated reliably, in an unambiguous fixed order with unambiguous start and end points.
There are many more sophisticated ways of synchronising, for example if the data is broken up into 'chunk's of 8 bits, and an odd parity bit is appended to each chunk, then the maximum '0's run-length is 16 bits. A sync pattern of 17 '0' bits could be used. Many variations are used, most of them involve limiting the maximum run-length of one binary value (or either binary value, when using differentially-coherent encoding) in the data section, and then using a longer run-length for the synchronisation.
Neck"lace (?; 48), n.
1.
A string of beads, etc., or any continuous band or chain, worn around the neck as an ornament.
2. Naut.
A rope or chain fitted around the masthead to hold hanging blocks for jibs and stays. |
FAQ
# How to make a trapezoid?
Best answer for this question, what makes a shape a trapezoid? A trapezoid is a quadrilateral with one pair of opposite sides parallel. It can have right angles (a right trapezoid), and it can have congruent sides (isosceles), but those are not required.
You asked, how do you draw a trapezoid?
As many you asked, what are two ways to make a trapezoid?
Likewise, how do you make a trapezoid with one square and one triangle?
It takes 3 triangles to make 1 trapezoid.
Contents
## What are the 3 types of trapezoid?
1. Right trapezoid – these trapezoids have a pair of right angles.
2. Isosceles trapezoid – trapezoids in which the non-parallel sides have the same length.
3. Scalene trapezoid – this type of trapezoid has four sides that are all of an unequal length.
## Can you draw a trapezoid with one right angle?
Explanation: A trapezoid can have either 2 right angles, or no right angles at all.
## What does a trapezoid look like?
A trapezoid is a four-sided flat shape with one pair of opposite parallel sides. It looks like a triangle that had its top sliced off parallel to the bottom. Usually, the trapezoid will be sitting with the longest side down, and you will have two sloping sides for the edges.
## What angles are in a trapezoid?
Like all other quadrangles, the sum of angles in a trapezoid is 360 degrees (or 2π radians). Since they have a pair of parallel sides, the trapezoid has an additional condition. The pair of angles along one of the legs are supplementary angles, which means their sum must be equal to 180 degrees (or π radians).
## What is a trapezoid 3rd grade?
A trapezoid is a quadrilateral with one pair of parallel sides. Two sides of a shape are parallel if lines placed along them never cross. Parallel. Not parallel. In a quadrilateral, parallel sides must be opposite sides.
## Is it possible to draw a trapezoid that is a rectangle?
Explanation: If a trapezoid is defined as a polygon with four sides (that is, quadrilateral) and two sides of it are parallel to each other, than a rectangle can be considered as a type of trapezoid. In this case all theorems proven for a trapezoid are true for rectangles.
## What is a trapezoid 1st grade?
A trapezoid, also known as a trapezium, is a flat closed shape having 4 straight sides, with one pair of parallel sides. The parallel sides of a trapezium are known as the bases, and its non-parallel sides are called legs.
## How do you make a trapezium with three triangles?
Work out an expression, in terms of b and h for the area of the trapezium. Here we have three identical triangles. Therefore the formula for the Area of the three triangles: A = (b x h) x 3 / 2 = A of trapezium.
## What do the midpoints of a trapezoid form?
When the midpoints of the two legs of a trapezoid are joined together, the resulting segment is called the median of the trapezoid.
## How many triangles does it take to make a hexagon?
A hexagon is made up of 6 congruent equilateral triangles. Each equilateral triangle has a length of 8 units.
## How a triangle and trapezoid related?
A trapezoid is lesser known than a triangle, but still a common shape. A trapezoid is a four-sided, two-dimensional shape with two parallel sides. Trapezoids have two bases. Those are the sides that are parallel.
## Can a square be a trapezoid?
Since a square has 4 sides of equal length, it can also be classified as a rhombus. The opposite sides are parallel so a square can also be classified as a parallelogram. If it is classified as a parallelogram then it is also classified as a trapezoid.
## How many corners does a trapezoid have?
A trapezoid has four corners. In 2-dimentional shapes, the point where two sides meet is called a corner, or a vertex. Since trapezoids are…
## What is a trapezoid with 2 right angles?
A right trapezoid (also called right-angled trapezoid) has two adjacent right angles.
See also How do i retrieve notes in my notebook hp pc with windows 10? |
# Elementary algebra: solving linear equations in one variable
Page 1 / 3
Elementary Algebra: An introduction to solving linear equations in one variable.
## Module overview
Learning how to solve various algebraic equations is one of our main goals in algebra. This module introduces the basic techniques for solving linear equations in one variable. (Prerequisites: Working knowledge of real numbers and their operations.)
## Objectives
• Define Linear Equations in One Variable
• Solutions to Linear Equations
• Solving Linear Equations
• Combining Like Terms and Simplifying
• Literal Equations
## Define linear equations in one variable
We begin by establishing some definitions.
Equation
An equation is a statement indicating that two algebraic expressions are equal.
Linear Equation in One Variable
A linear equation in one variable $x$ is an equation that can be written in the form $\text{ax}+b=0$ where $a$ and $b$ are real numbers and $a\ne 0$ .
Following are some examples of linear equations in one variable, all of which will be solved in the course of this module.
$x+3=-5$
$\frac{x}{3}+\frac{1}{2}=\frac{2}{3}$
$5\left(3x+2\right)-2=-2\left(1-7x\right)$
## Solutions to linear equations in one variable
The variable in the linear equation $2x+3=\text{13}$ is $x$ . Values that can replace the variable to make a true statement compose the solution set. Linear equations have at most one solution. After some thought, you might deduce that $x=5$ is a solution to $2x+3=\text{13}$ . To verify this we substitute the value 5 in for $x$ and see that we get a true statement, $2\left(\mathbf{5}\right)+3=\text{10}+3=\text{13}$ .
Is $x=3$ a solution to $-2x-3=-9$ ?
Yes, because $-2\left({3}\right)-3=-6-3=-9$
Is $a=-\frac{1}{2}$ a solution to $-\text{10}a+5=\text{25}$ ?
No, because $-\text{10}\left({-}\frac{{1}}{{2}}\right)+5=5+5=\text{10}\ne \text{25}$
When evaluating expressions, it is a good practice to replace all variables with parenthesis first, then substitute in the appropriate values. By making use of parenthesis we could avoid some common errors using the order of operations.
Is $y=-3$ a solution to $2y-5=-y-\text{14}$ ? Yes because $y=-3$ produces a true mathematical statement.
## Solving linear equations in one variable
When the coefficients of linear equations are numbers other than nice easy integers, guessing at solutions becomes an unreasonable prospect. We begin to develop an algebraic technique for solving by first looking at the properties of equality.
## Properties of equality
Given algebraic expressions A and B where c is a real number:
## Division property of equality
Multiplying or dividing both sides of an equation by zero is carefully avoided. Dividing by zero is undefined and multiplying both sides by zero will result in an equation 0=0.
To summarize, the equality is retained if we add, subtract, multiply and divide both sides of an equation by any nonzero real number. The central technique for solving linear equations involves applying these properties in order to isolate the variable on one side of the equation.
Use the properties of equality to solve: $x+3=-5$ The solution set is $\left\{-8\right\}$ .
#### Questions & Answers
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
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Berger describes sociologists as concerned with
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### UnexpectedValue's blog
By UnexpectedValue, 7 months ago,
I could not find any article or blog on this idea, and felt having one could be helpful to many. There might be unwanted errors, so please feel free to to share any suggestions, corrections or concerns. For those who are unware of Binary Exponentation, this CP-Algorithms article or this video by Errichto can help get familiar with it.
#### Introduction
Now that we know the idea behind binary exponentiation; let us try expanding the idea. Say, we break the problem into two parts, $L$ and $R$, such that, once we have both the results, combining them gives us our final answer as $ans = L \odot R$, where $\odot$ is an associative binary operator. If we can find $R$ as a function of $L$ as in $R = f(L)$ fast enough, say $O(X)$, then our final answer can be found in $O(X\;log(n))$. The idea is simple enough, so we now have a look at some simple examples. For the sake of simplicity we assume problem size $n$ is of form $2^k$ , $k \in \mathbb{Z}$ .
#### Example 1: Calculating $a^n$
We are starting from the simplest example. We know that $ans = \underbrace{a \cdot a \dots \cdot a \cdot a }_\text{ n times}$. If we take $R = f(L) = L$, where $L = a^{n/2}$; we can find $R$ in $O(1)$ and build our answer as -
$\newline$
$ans = \underbrace{\underbrace{\dots \dots \dots}_{L} \odot \underbrace{\dots \dots \dots}_{R=f(L)}}_\text{L'} \odot \underbrace{\dots \dots \dots \dots \dots \dots \dots}_\text{R' = f(L')} \; \odot \dots \dots$
We are able to compute $R'$ directly from $L'$, unlike other methods such as RMQ or Binary Lifting where we build our answer from already calculated $L$ and $R$ on the independent smaller ranges to later combine them to compute the the answer for the bigger range.
Time Complexity: Finding $R = f(L)$ is $O(1)$, so total time complexity is $O(log(n))$.
Implementation
#### Example 2: Summation of GP Series
Considering a standard GP series, we find its summation as
$\newline$
$ans = 1 + r + r^{2} \dots r^{n-1} = \frac{r^{n}-1}{r-1}$
If we have to find the summation under arbitrary MOD then it's not necessary that $(r-1)^{-1}$ exists, and so we alternatively express our sum as
$ans = \underbrace{1 + r + r^{2} \dots r^{ \frac{n-1}{2} } }_\text{L} + \underbrace{r^{ \frac{n}{2} } + \dots + r^{n-1}}_\text{R}$
which gives $R = f(L) = r^{\frac{n}{2}} \cdot L$.
To handle the case for odd $n$ during implementation, we can break series as
$\newline$
$ans = 1 + a \cdot \underbrace{(1 + a + a^2 + \dots + a^{n-2})}_\text{say sum is S}$
We can find $S$ as $(n - 1)$ is even, and final answer will be $= 1 + a \cdot S$.
Time Complexity: Computing $R = f(L)$ is $O(log(n))$, so our total time complexity becomes $O(log(n)^{2})$. But we can calculate powers of $r$ along with our series, so we restrict the final time complexity to $O(log(n))$.
Implementation
We can use a similar idea for calculating GP series of matrices $= I + A + A^2 + \dots + A^{n-1}$. Just like integers under modulo, we can't always guarantee $(A-I)^{-1}$ exists.
Contest Example: In AtCoder ABC 293 Task E the task is to just compute the sum of a GP series.
#### Example 3: Summation of AP-GP series
We start by considering the simple series $\displaystyle r + 2 \cdot r^2 + 3 \cdot r^3 \dots + (n-1)\cdot r^{n-1} = \sum\limits_{i=0}^{n-1} i\cdot r^i$. For this we express the summation as below
$ans = \underbrace{r + 2 \cdot r^2 + \dots + (\frac{n}{2}-1) \cdot r^{ \frac{n}{2}-1 }}_\text{L} + \underbrace{ \frac{n}{2} \cdot r^{\frac{n}{2}} + \dots + (n-1) \cdot r^{n-1}}_\text{R}$
Now we solve for $R = f(L)$.
Solution for R as a function of L
To handle the odd case during implementation, we express summation as:
$\newline$
$sum = r + 2\cdot r^2 + \dots + (n-1) \cdot r^{n-1}$
$= \left(r + r^2 + \dots + r^i + \dots + r^{n-1} \right) + \left( r^2 + 2\cdot r^3 + \dots + (i-1)\cdot r^{i-1} + \dots + (n-2)\cdot r^{n-1} \right)$
$= r \cdot \underbrace{\left(1 + r^1 + \dots + r^i + \dots + r^{n-2} \right)}_\text{X} + r \cdot \underbrace{\left( r^1 + 2\cdot r^2 + \dots + i\cdot r^{i} + \dots + (n-2)\cdot r^{n-2} \right)}_\text{Y}$
We can get $X$ and $Y$ because they are of even length, and our final answer will be $= r \cdot X + r \cdot Y$ .
Time Complexity: To get $R$ we need to solve $L_{0}$ which takes $O(log(n))$ so total will be $O(log(n)^2)$, but we can calculate powers of $r$, $L_{0}$ and $L$ together which makes total time $O(log(n))$.
Implementation
For any generic AP-GP series $\displaystyle = \sum\limits_{i=0}^{n-1} ( a + i \cdot b ) \cdot r^i$ $= a \cdot \underbrace{\sum\limits_{i=0}^{n-1} r^i}_\text{ L0 } + b \cdot \underbrace{\sum\limits_{i=0}^{n-1} i \cdot r^i}_\text{L} = a \cdot L_{0} + b \cdot L$, where both are being calculated with single function call.
AtCoder ABC 129 task F is a good practice example for the ideas in the blog.
#### Genralising the above series summations
Now, let us consider $\displaystyle S(n,m) = \sum\limits_{i=0}^{n-1} i^m \cdot r^i$. Then we would have GP $= S(n,0)$ and AP-GP $= S(n,1)$. Hence,
$\newline$
$\displaystyle S(n,m) = r + 2^m \cdot r^2 + \dots + i^m \cdot r^i + \dots + (n-1)^m \cdot r^{n-1}$
$\displaystyle = \underbrace{r + 2^m \cdot r^2 + \dots + (\frac{n}{2}-1)^{m} \cdot r^{\frac{n}{2}-1}}_\text{L(m,n/2)} + \underbrace{{(\frac{n}{2}})^{m} \cdot r^{\frac{n}{2}} + \dots + (n-1)^m \cdot r^{n-1}}_\text{R(m,n/2)}$
Here $L_{m,\frac{n}{2}} = S(\frac{n}{2},m)$. Then, $S(n,m) = L_{m,\frac{n}{2}} + R_{m,\frac{n}{2}}$, We need to compute $R_{m,\frac{n}{2}}$.
$\newline$
$\displaystyle R_{m,\frac{n}{2}} = r^{\frac{n}{2}} \cdot \left( (\frac{n}{2})^m + (\frac{n}{2}+1)^m \cdot r + \dots + (\frac{n}{2} + i)^m \cdot r^i + \dots (\frac{n}{2} + \frac{n}{2} - 1)^m \cdot r^{\frac{n}{2}-1} \right)$
$\displaystyle R_{m,\frac{n}{2}} = r^{\frac{n}{2}} \cdot \left( \sum\limits_{i=0}^{\frac{n}{2}-1} \left( \sum\limits_{j=0}^{m} \binom{m}{j} \left(\frac{n}{2}\right)^{m-j} \cdot i^j \right) \cdot r^i \right)$
Now, swapping the summations -
$\displaystyle R_{m,\frac{n}{2}} = r^{\frac{n}{2}} \cdot \left( \sum\limits_{j=0}^{m} \binom{m}{j} \left( \frac{n}{2}\right) ^{m-j} \underbrace{\left( \sum\limits_{i=0}^{\frac{n}{2}-1} i^j \cdot r^i \right)}_\text{L(j,n/2)} \right)$
Finally,
$\displaystyle R_{m,\frac{n}{2}} = r^{\frac{n}{2}} \cdot \left( \sum\limits_{j=0}^{m} \binom{m}{j} \left( \frac{n}{2}\right) ^{m-j} \cdot L_{j,\frac{n}{2}} \right) = r^{\frac{n}{2}} \cdot \left( \sum\limits_{j=0}^{m} \binom{m}{j} \left( \frac{n}{2}\right) ^{m-j} \cdot S_{\frac{n}{2},j} \right)$
As $j \leq m$ if we calculate everything together, just like above and hence know the value of every $L_{j,\frac{n}{2}}$ without any extra time. We can validate our results for the above illustrated examples of AP and AP-GP series.
Validating our results against our previous examples.
We can now evaluate our original summation as $\displaystyle S_{n,m} = S_{\frac{n}{2},m} + r^{\frac{n}{2}} \cdot \left( \sum\limits_{j=0}^{m} \binom{m}{j} \left( \frac{n}{2}\right) ^{m-j} \cdot S_{\frac{n}{2},j} \right)$.
##### Handling Odd Case
We have $\displaystyle S_{n,m} = \sum\limits_{i=0}^{n-1} i^m \cdot r^i = r \cdot \left( \sum\limits_{i=1}^{n-2} (i+1)^m \cdot r^i \right)$. However we can only write this if $m>0$ but for $m=0$ first term will not be $0$ but instead be $\displaystyle \lim_{x \to 0} x^x = 1$. So for $m=0$ we will add $1$ instead of $0$, keeping the remainder of the expression same. Thus
$\displaystyle S_{n,m} = r \cdot \left( \sum\limits_{i=0}^{n-2} \sum\limits_{j=0}^{m} \binom{m}{j} i^j \cdot r^i \right)$
And swapping summations
$\displaystyle S_{n,m} = r \cdot \left( \sum\limits_{j=0}^{m} \binom{m}{j} \underbrace{\sum\limits_{i=0}^{n-2} i^j \cdot r^i}_\text{S(n-1,j)} \right) = r \cdot \left( \sum\limits_{j=0}^{m} \binom{m}{j} S_{n-1,j} \right) \hspace{0.25cm} \text{for} \hspace{0.25cm}m > 0$
Hence, we have the final result as:
$\displaystyle $$S_{n,m}= \begin{cases} 1 + r \cdot S_{n-1,0} & \text{if } m=0\\ r \cdot \left(\displaystyle \sum\limits_{j=0}^{m} \binom{m}{j} S_{n-1,j} \right) & \text{if } m > 0 \end{cases}$$$
We can also construct a clean visualization of the even and odd cases using matrix multiplication instead of summation. These are illustrated as below.
Odd Case.
Even Case.
In the above cases, whenever we are computing values of $R_{j, n / 2} \hspace{0.25cm} \text{for} \hspace{0.25cm} j \in [0, m]$, they still indeed maintain our original form of $R = f(L)$ as we can express them as
$\vec{R_{\frac{n}{2}}} = f(\vec{L})= r^{\frac{n}{2}} \left( M(n,m) \cdot \vec{L_{\frac{n}{2}}} \right)$
Time Complexity: To find $R_{j,\frac{n}{2}}$ takes summation of $m$ terms and we do that for every $j \leq m$ which will take $O(m^2)$. So, total complexity is $O(m^2 \cdot log(n)).$
Implementation
• +153
» 7 months ago, # | -14 Orz!
» 7 months ago, # | +1 Nice blog!btw S(n, m) can be calculated in subquadratic time. For more information, see the Forum section of this Library Checker problem.
• » » 7 months ago, # ^ | 0 Thanks for sharing this information. This is new to me and I will certainly have a look into it. Seems intriguing.
• » » 7 months ago, # ^ | 0 In the forum we are given a prime $MOD (= 998244353)$ and the solution in the forum is using that fact too, but how do we find summation for arbitrary $MOD$ in subquadratic time, where we are not guaranteed modulo inverse exists?
» 7 months ago, # | ← Rev. 5 → +26 Example 2 even case can be much easier:$a^0 + a^1 + \cdots + a^{n-1}$ = $a^0 + a^2 + a^4 + \cdots + a^1 + a^3 + a^5 + \cdots$ = $(a + 1) \times (a^0 + a^2 + \cdots + a^{n-2})$ codetemplate T sg(T a, ll n) { if(n == 0) return 0; if(n%2) return sg(a, n-1) * a + 1; return sg(a*a, n / 2) * (a + 1); }
• » » 7 months ago, # ^ | 0 Thanks for adding this easy to implement idea to the one shared in the blog! Will certainly come in handy. |
# How many ways can 10 similar beads be arranged to form a necklace?
Contents
Answer: This is called a cyclic permutation. The formula for this is simply (n-1)!/2, since all the beads are identical. Hence, the answer is 9!/2 = 362880/2 = 181440.
## How many necklaces can be made by using 10 round beads all of a different colors?
There are 10 beads of distinct colours; say, A, B, C, D, E, F, G, H, I and J. If no restriction is imposed then, there are (10!) = 3628800 ways to put these ten distinctly coloured beads into a necklace.
## How many ways can 5 different beads be arranged to form a necklace?
So, we have to divide 24 by 2. Therefore the total number of different ways of arranging 5 beads is 242=12 .
IT IS INTERESTING: How do you get Diamond Skin in Cold War?
2520. 5040.
## How many necklaces can be made using 7 beads of which 5 are identical red beads and 2 are identical blue beads?
= 720/(120*2) = 3. So we can have 3 different necklaces.
## How many necklaces can be formed with 6 white and 5 red beads if each necklace is unique how many can be formed?
5! but correct answer is 21.
## How many ways can you make a bracelet with 5 different beads?
Thus for n=5, there are possible 4!/2=12 different bracelets.
## How many different change can be made using 5 different Coloured beads?
So there can be 12 different arrangements.
## How many ways can we arrange 5 different colors in a circular arrangement?
Example 7.4.
We are looking for permutations for the letters HHHHTT. The answer is 6! 4! 2!
## How many necklaces of 12 beads each can be made from 18 beads of various Colours?
Correct Option: C
First, we can select 12 beads out of 18 beads in 18C12 ways. Now, these 12 beads can make a necklace in 11! / 2 ways as clockwise and anti-clockwise arrangements are same. So, required number of ways = [ 18C12 . 11! ] / 2!
## How many ways can 8 beads of different Colour be strung on a ring?
2520 Ways 8 beads of different colours be strung as a necklace if can be wear from both side.
## How many ways can a necklace be formed from 2 red and 2 blue beads?
= 1,680. Total of permutations = 2,520+3*1,680 = 7,560. |
# What is the range equation?
The Range Equation is R = vi. 2 sin 2θi.
## How do you find the range in physics?
Maximum Range of Projectile Now that the range of projectile is given by R = u 2 sin 2 θ g , when would be maximum for a given initial velocity .
## What is a range in physics?
Assuming a projectile is launched from the ground level, the range is defined as the distance between the launch point and the point where the projectile hits the ground.
## How do you use the range formula?
In order to calculate the range using the range formula, there are two mean steps to be followed: Step 1: Place all the numbers from the lowest to the highest value in the data set. Step 2: By using the range formula, subtract the lowest value from the highest value picked from the data set.
## What is range Value?
The range is the simplest measurement of the difference between values in a data set. To find the range, simply subtract the lowest value from the greatest value, ignoring the others.
## What is the range of a projectile in physics?
The range of the projectile is the displacement in the horizontal direction. There is no acceleration in this direction since gravity only acts vertically. shows the line of range. Like time of flight and maximum height, the range of the projectile is a function of initial speed.
## What is the correct example of a range?
The Range is the difference between the lowest and highest values. Example: In 4, 6, 9, 3, 7 the lowest value is 3, and the highest is 9. So the range is 9 − 3 = 6.
## Is range the same as distance physics?
The distance traveled horizontally from the launch position to the landing position is known as the range. The range of an angled-launch projectile depends upon the launch speed and the launch angle (angle between the launch direction and the horizontal).
## How do you find the maximum height and range of a projectile?
hmax = h + V₀² / (4 * g) and in that case, the range is maximal if launching from the ground (h = 0). if α = 0°, then vertical velocity is equal to 0 (Vy = 0), and that’s the case of horizontal projectile motion.
## What is the horizontal range of a projectile if 90?
As we know that sinθ is maximum at 90°. Therefore horizontal range will maximum at 45°.
## How do you find the range of a frequency distribution?
In case of continuous frequency distribution, range, according to the definition, is calculated as the difference between the lower limit of the minimum interval and upper limit of the maximum interval of the grouped data. That is for X: 0-10, 10-20, 20-30 and 40-50, range is calculated as 40-0=40.
## What is range and mean?
Mean is the average of all of the numbers. Median is the middle number, when in order. Mode is the most common number. Range is the largest number minus the smallest number.
## How do you find the range and standard deviation?
The standard deviation is approximately equal to the range of the data divided by 4. That’s it, simple. Find the largest value, the maximum and subtract the smallest value, the minimum, to find the range. Then divide the range by four.
## What is the range of the projectile R?
The range (R) of the projectile is the horizontal distance it travels during the motion. Using this equation vertically, we have that a = -g (the acceleration due to gravity) and the initial velocity in the vertical direction is usina (by resolving). Hence: y = utsina – ½ gt2 (1)
## How do you find the range of a horizontal projectile?
Range of the projectile r = V * t = v * √(2 * h / g) . We won’t calculate the maximum height here, as we don’t have an initial vertical velocity component – and that means that the maximal height is the one from which we’re starting.
## What are the uses of range?
Applications of Range The range is an easy but good way to get a basic understanding of the spread of the numbers in the data set. Moreover, it is easy to calculate as it only requires a basic arithmetic operation. The range can also be used to estimate another measure of spread which is the standard deviation.
## How do you find the range of a negative number?
The lowest or coldest temperature is negative 10 degrees Fahrenheit. We can, therefore, calculate the range by subtracting negative 10 from 12. Subtracting negative 10 from a number is the same as adding 10 to that number. 12 plus 10 is equal to 22.
## How do I calculate the median?
1. Step 1: Arrange the scores in numerical order.
2. Step 2: Count how many scores you have.
3. Step 3: Divide the total scores by 2.
4. Step 4: If you have an odd number of total scores, round up to get the position of the median number.
## What is projectile trajectory and range?
The object’s maximum height is the highest vertical position along its trajectory. The horizontal displacement of the projectile is called the range of the projectile. The range of the projectile depends on the object’s initial velocity.
## Why is Max range at 45 degrees?
The sine function reaches its largest output value, 1, with an input angle of 90 degrees, so we can see that for the longest-range punts 2θ = 90 degrees and, therefore, θ = 45 degrees. A projectile, in other words, travels the farthest when it is launched at an angle of 45 degrees.
## What is the maximum range?
The greatest distance a weapon can fire without consideration of dispersion.
## What is the symbol of range?
R indicates range. When using set notation, inequality symbols such as ≥ are used to describe the domain and range. |
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# NCERT Solutions for Class 10 Maths Chapter 8- Introduction to Trigonometry: Download Free PDF
Trigonometry is one the most important fundamental topics in Class 8 Maths. Trigonometry, which comes from a Greek word in which 'trigono' means 'triangle' and 'metry' means 'to measure', is an essential branch of mathematical studies. It deals with the relationship between side lengths and angles of triangles.
Trigonometry's uses are vast, as its applications are enormous worldwide. In this topic, triangles play a vital role, such as triangles with one angle equal to 90 degrees, right-angled triangles to be more precise. It helps us find missing angles of triangles as well as the missing sides and their lengths.
The chapter explores the below-mentioned trigonometric formulae:
Let us consider the three points of a right-angled triangle as OAB, where OA & OB makes 90° to each other,
Where,
OB is the opposite side
AB is the hypotenuse side
The hypotenuse side of a triangle is always the lengthiest one, whereas the other two sides are comparatively shorter.
So, from the given information, this formula is derived,
sin = opposite / hypotenuse
The NCERT Maths Chapter 8 Trigonometry has the following exercises; Firstly, it deals with exercise problems that can be pictured using the right angle triangle. Secondly, it consists of an introduction on trigonometric ratios and their examples, and practice problems can accompany this consisting of the derivation of sine, cosine, tangent, and other trigonometric functions.
In addition to this, the third part consists of measurements related to trigonometric ratios, while the fourth one deals with few solved problems and trigonometric ratio criteria for complementary angles with an exercise. Finally, the last section is all about subjects related to identities of trigonometry and its example problems. So, this is the overall overview of this chapter. Provided below is the detailed video explanation for the chapter.
## FAQs for Class 10 Maths NCERT Solutions
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A. NCERT Solutions for Class 10 Maths are designed for the NCERT books, and they follow the CBSE and NCERT guidelines; thus, they may not be according to the guidelines and specifications of other boards, but the students can still gain the same knowledge on the concepts and the topics that have been discussed in the NCERT books. Since the majority of the syllabus is similar across all boards in the subject of Maths thus, the NCERT Solutions for Class 10 Maths prove beneficial to the students of all boards
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A. The NCERT Solutions for Class 10 Maths is available at the website of the Aakash Institute. The PDF of the solutions is available free of cost, and it is available to everyone regardless of the board, you are a student of Aakash institute or not.
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A. The NCERT solutions for class 10 Maths by Aakash specifically provide detailed answers for all subjects and step-by-step solutions for Maths. Accessing the NCERT Solutions by Aakash can give you an edge over others. Moreover, it is based on the exact syllabus which covers not only the CBSE board but also the Olympiads.
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A. For the ease of students, we at Aakash have provided a PDF file for NCERT Solutions 2020 of all subjects, all topics for all classes. The detailed solutions are prepared by our senior and experienced faculty which gives students an in-depth understanding of the concepts rather than the particular question.
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A. Well, it’s a fact that NCERT books are the best for CBSE board exam preparation. It is not just because this council is recommended by CBSE itself but the content and syllabus of NCERT books are the best across the country. Thus, it gets extremely important for students to thoroughly go through the recommended NCERT syllabus for a particular exam. |
# 4.7.1: Determining the Equation of a Line
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## Determining the Equation of a Line
You are probably aware that many real-world situations can be described with linear graphs and equations. In this lesson, we’ll see how to find equations in a variety of situations.
#### Write an Equation Given Slope and y−Intercept
Recall that you may write an equation in slope–intercept form with a few simple steps: start with the general equation for the slope-intercept form of a line, y=mx+b, and then substitute the given values of m and b into the equation. For example, a line with a slope of 4 and a y−intercept of –3 would have the equation y=4x−3.
If you are given just the graph of a line, you can identify the slope and y−intercept from the graph and write the equation from there. For example, on the graph below you can see that the line rises by 1 unit as it moves 2 units to the right, so its slope is 1/2. Also, you can see that the y−intercept is –2, so the equation of the line is y=1/2x−2.
#### Write an Equation Given the Slope and a Point
Often, we don’t know the value of the y−intercept, but we know the value of y for a non-zero value of x. In this case, it’s often easier to write an equation of the line in point-slope form. An equation in point-slope form is written as y−y0=m(x−x0), where m is the slope and (x0,y0) is a point on the line.
#### Writing the Equation of a Line in Point-Slope Form
A line has a slope of 3/5, and the point (2, 6) is on the line. Write the equation of the line in point-slope form.
Plug in 3/5 for m, 2 for x0 and 6 for y0.
The equation in point-slope form is y−6=3/5(x−2).
Notice that the equation in point-slope form is not solved for y. If we did solve it for y, we’d have it in y−intercept form. To do that, we would just need to distribute the 35 and add 6 to both sides. That means that the equation of this line in slope-intercept form is y=3/5x−(6/5)+6, or simply y=(3/5)x+24/5.
#### Write an Equation Given Two Points
Point-slope form also comes in useful when we need to find an equation given just two points on a line.
For example, suppose we are told that the line passes through the points (-2, 3) and (5, 2). To find the equation of the line, we can start by finding the slope.
Starting with the slope formula, m=(y2−y1)/(x2−x1), we plug in the x− and y−values of the two points to get m=(2−3)/(5−(−2))=−1/7. We can plug that value of m into the point-slope formula to get y−y0=-1/17(x−x0).
Now we just need to pick one of the two points to plug into the formula. Let’s use (5, 2); that gives us y−2=−1/7(x−5).
What if we’d picked the other point instead? Then we’d have ended up with the equation y−3=−1/7(x+2), which doesn’t look the same. That’s because there’s more than one way to write an equation for a given line in point-slope form. But let’s see what happens if we solve each of those equations for y.
Starting with y−2=−1/7(x−5), we distribute the −1/7 and add 2 to both sides. That gives us y=−1/7x+(5/7)+2, or y=−1/7x+19/7.
On the other hand, if we start with y−3=−1/7(x+2), we need to distribute the −1/7 and add 3 to both sides. That gives us y=−1/7x−(2/7)+3, which also simplifies to y=−1/7x+19/7.
So whichever point we choose to get an equation in point-slope form, the equation is still mathematically the same, and we can see this when we convert it to y−intercept form.
#### Writing an Equation in y−intercept form
A line contains the points (3, 2) and (-2, 4). Write an equation for the line in point-slope form; then write an equation in y−intercept form.
Find the slope of the line: m=(y2−y1)/(x2−x1)=(4−2)/(−2−3)=−2/5
Plug in the value of the slope: y−y0=−2/5(x−x0).
Plug point (3, 2) into the equation: y−2=−2/5(x−3).
The equation in point-slope form is y−2=−2/5(x−3).
To convert to y−intercept form, simply solve for y:
y−2=−2/5(x−3)→y−2=−2/5x+6/5→y=−2/5x+(6/5)+2→y=−2/5x+31/5.
The equation in y−intercept form is y=−2/5x+31/5.
#### Graph an Equation in Point-Slope Form
Another useful thing about point-slope form is that you can use it to graph an equation without having to convert it to slope-intercept form. From the equation y−y0=m(x−x0), you can just read off the slope m and the point (x0,y0). To draw the graph, all you have to do is plot the point, and then use the slope to figure out how many units up and over you should move to find another point on the line.
#### Graphing an Equation of a Line
Make a graph of the line given by the equation y+2=2/3(x−2).
To read off the right values, we need to rewrite the equation slightly: y−(−2)=2/3(x−2). Now we see that point (2, -2) is on the line and that the slope is 23.
First plot point (2, -2) on the graph:
A slope of 23 tells you that from that point you should move 2 units up and 3 units to the right and draw another point:
Now draw a line through the two points and extend it in both directions:
### Examples
Example 4.7.1.1
A line contains the points (1, -2) and (0, 0). Write an equation for the line in point-slope form; then write an equation in y−intercept form.
Solution
Find the slope of the line: m=(y2−y1)/(x2−x1)=(−2−0)/(1−0)=−2/1=−2
Plug in the value of the slope: y−y0=-2(x−x0).
Plug point (1, -2) into the equation: y−(−2)=−2(x−1).
The equation in point-slope form is y+2=−2(x−1).
To convert to y−intercept form, simply solve for y:
y+2=−2(x−1)→y+2=−2x+2→y=−2x+2−2→y=−2x.
The equation in y−intercept form is y=−2x.
### Review
Find the equation of each line in slope–intercept form.
1. The line has a slope of 7 and a y−intercept of -2.
2. The line has a slope of -5 and a y−intercept of 6.
3. The line has a slope of −1/4 and contains the point (4, -1).
4. The line contains points (3, 5) and (-3, 0).
5. The line contains points (10, 15) and (12, 20).
Write the equation of each line in slope-intercept form.
Find the equation of each linear function in slope–intercept form.
1. m=5,f(0)=−3
2. m=−7,f(2)=−1
3. m=1/3,f(−1)=2/3
4. m=4.2,f(−3)=7.1
5. f(14)=3/4,f(0)=5/4
6. f(1.5)=−3,f(−1)=2
Write the equation of each line in point-slope form.
1. The line has slope −1/10 and goes through the point (10, 2).
2. The line has slope -75 and goes through the point (0, 125).
3. The line has slope 10 and goes through the point (8, -2).
4. The line goes through the points (-2, 3) and (-1, -2).
5. The line contains the points (10, 12) and (5, 25).
6. The line goes through the points (2, 3) and (0, 3).
7. The line has a slope of 3/5 and a y−intercept of -3.
8. The line has a slope of -6 and a y−intercept of 0.5.
Write the equation of each linear function in point-slope form.
1. m=−1/5 and f(0)=7
2. m=−12 and f(−2)=5
3. f(−7)=5 and f(3)=−4
4. f(6)=0 and f(0)=6
5. m=3 and f(2)=−9
6. m=−9/5 and f(0)=32 |
# NCERT Solutions for Class 8 Maths
Find 100% accurate solutions for NCERT Class VIII Math. All questions have been explained and solved step-by-step, to help you understand thoroughly. Free Download option available!
1.
2.
3.
4. Solution:
5. (i) No, a polyhedron cannot have 3 triangles for its faces.
(ii) Yes, a polyhedron can have four triangles which is known as pyramid on triangular
(iii) Yes, a polyhedron has its faces a square and four triangles which makes a pyramid on square base.
6. It is possible, only if the number of faces are greater than or equal to 4.
7. Figure (ii) unsharpened pencil and figure (iv) a box are prisms.
8. (i) A prism becomes a cylinder as the number of sides of its base becomes larger and larger.
(ii) A pyramid becomes a cone as the number of sides of its base becomes larger and larger.
9. No, it can be a cuboid also.
10. (i) Here, figure (i) contains 7 faces, 10 vertices and 15 edges.
Using Eucler’s formula, we see F + V – E = 2
Putting F = 7, V = 10 and E = 15,
F + V – E = 2 ⇒ 7 + 10 – 5 = 2
⇒ 17 – 15 = 2 ⇒ 2 = 2
⇒ L.H.S. = R.H.S.
(ii) Here, figure (ii) contains 9 faces, 9 vertices and 16 edges.
Using Eucler’s formula, we see F + V – E = 2
F + V – E = 2 ⇒ 9 + 9 – 16 = 2
⇒ 18 – 16 = 2 ⇒ 2 = 2
⇒ L.H.S. = R.H.S.
Hence verified Eucler’s formula.
11. In first column, F = ?, V = 6 and E = 12
Using Eucler’s formula, we see F + V – E = 2
F + V – E = 2 ⇒ F + 6 – 12 = 2
⇒ F – 6 = 2 ⇒ F = 2 + 6 = 8
Hence there are 8 faces.
In second column, F = 5, V = ? and E = 9
Using Eucler’s formula, we see F + V – E = 2
F + V – E = 2 ⇒ 5 + V – 9 = 2
⇒ V – 4 = 2 ⇒ V= 2 + 4 = 6
Hence there are 6 vertices.
In third column, F = 20, V = 12 and E = ?
Using Eucler’s formula, we see F + V – E = 2
F + V – E = 2 ⇒ 20 + 12 – E = 2
⇒ 32 – E = 2 ⇒ E = 32 – 2 = 30
Hence there are 30 edges.
12. If F = 10, V = 15 and E = 20.
Then, we know Using Eucler’s formula, F + V – E = 2
L.H.S. = F + V – E
= 10 + 15 – 20
= 25 – 20
=5
R.H.S. = 2
∵ L.H.S. ≠ R.H.S.
Therefore, it does not follow Eucler’s formula.
So polyhedron cannot have 10 faces, 20 edges and 15 vertices.
MySchoolPage connects you with exceptional, certified math tutors who help you stay focused, understand concepts better and score well in exams! |
# Lesson (II)
## Families of Sets
We might be interested at time to consider not one set but a bunch of them at the same time.
For example, we might be interested in a sequence
$A_0, A_1, A_2, \ldots$ of sets indexed by the natural numbers. Here the numbers $0,1,2,\ldots$ are the indices, whereas the sets $A_n$ might or might not have any natural numbers in them: For each natural number $n$, we can define the set $A_n$ to be the set of people alive today that are of age $n$.
More generally, if $I$ is a set, we sometimes wish to consider a family $$\set{ A_i \mid i \in I }$$ of sets indexed by elements of $I$. Here the member of the set $$\set{ A_i \mid i \in I }$$ are themselves sets, indexed by elements of $I$.
#### Union and intersection of indexed families
Given a family $\set{ A_i \mid i \in I }$ of sets indexed by $I$, we can form its union: $\bigcup_{i \in I} A_i = \set{ x \mid x \in A_i \text{ for some i \in I} }$ We can also form the intersection of this family: $\bigcap_{i \in I} A_i = \set{ x \mid x \in A_i \text{ for every i \in I} }$
So an element $x$ is in $\bigcup_{i \in I} A_i$ if and only if $x$ is in $A_i$ for some $i$ in $I$, and $x$ is in $\bigcap_{i \in I} A_i$ if and only if $x$ is in $A_i$ for every $i$ in $I$.
The operations of union and intersection of families of sets are represented in symbolic logic by the existential and the universal quantifiers. We have: $\forall x \; (x \in \bigcup_{i \in I} A_i \iff \exists i \in I \; (x \in A_i))$ $\forall x \; (x \in \bigcap_{i \in I} A_i \iff \forall i \in I \; (x \in A_i))$
#### Challenge
The operations of indexed union and intersection have a particularly simple nature when the indexing set has finite number of elements:
1. Describe these operations when indexing set is empty.
2. Describe these operations when indexing set is a singleton.
3. Describe these operations when indexing set has exactly two elements.
#### Challenge (Disjoint Unions for Families of Sets)
We defined the disjoint union of two sets by $A \times \set{0} \cup B \times \set{1}$. Can you extend this definition for a family of sets $\set{A_i \mid i \in I}$? (Note that by extension here we mean that your definition of the disjoint union of a family of sets should agree with the usual disjoint union of sets when the family has only two members.)
The following theorem states that the usual intersection and union distributes over indexed unions and intersections:
Theorem. Let $\set{ B_i \mid i \in I }$ be a family of sets. We have
1. $$A \cap \bigcup_{i \in I} B_i = \bigcup_{i \in I} (A \cap B_i)$$
2. $$A \cup \bigcap_{i \in I} B_i = \bigcap_{i \in I} (A \cup B_i)$$
#### Challenge
Prove the theorem above.
We can have a family of sets indexed by many sets: for instance, a family $\set{A_{i,j} \mid i\in I, j \in J}$.
For every such family, consider the family $\set{ B_i \mid i \in I}$ where $B_i = \bigcup_{j \in J} A_{i,j}$ (here we fix $i \in I$, and let $j$ range over $J$).
We define $\bigcup_{i \in I} \bigcup_{j \in J} A_{i,j}$ to be $\bigcup_{i \in I} B_i$.
#### Challenge
Prove the following equalities of sets:
1. $\bigcup_{i \in I} \bigcup_{j \in J} A_{i,j} = \bigcup_{j \in J} \bigcup_{i \in I} A_{i,j}$
2. $\bigcap_{i \in I} \bigcap_{j \in J} A_{i,j} = \bigcap_{j \in J} \bigcap_{i \in I} A_{i,j}$.
#### Challenge
1. Prove that for every family $\set{ A_{i,j} \mid i \in I, j \in J}$ of sets we have $\bigcup_{i \in I} \bigcap_{j \in J} A_{i,j} \subseteq \bigcap_{j \in J} \bigcup_{i \in I} A_{i,j}$
2. Find the indexing sets $I$ and $J$ and a family $\set{ A_{i,j} \mid i \in I, j \in J}$ such that $\bigcap_{j \in J} \bigcup_{i \in I} A_{i,j} \nsubseteq \bigcup_{i \in I} \bigcap_{j \in J} A_{i,j}$
## Power Sets
Let $X$ be a set. The power set of $X$, written $\mathcal{P}(X)$ is the set of all subsets of $X$. Formally, $\mathcal{P}(X) \defeq \set{ U \mid U \subseteq X }$ Note that the predicate we use to form the set above is $Sub_X$, that is $Sub_X (U)$ is the sentence that $U$ is a subset of $X$.
Therefore, $\forall U \, \big( U \subseteq X \Leftrightarrow U \in \mathcal{P}(X) \big)$ Note that the power set of every set is inhabited since for a set $X$ we have $\varnothing \in \mathcal{P}(X)$ and $X \in \mathcal{P}(X)$.
#### Challenge
Suppose $A$ is a set with $3$ elements. How many elements does the set $\mathcal{P}\set{\emptyset,A}$ have?
With the definition of power set, let us observe that
• $\mathcal{P} \emptyset = \set{\emptyset}$
• $\mathcal{P}\mathcal{P} \emptyset= \set{ \emptyset, \set{\emptyset} }$
#### Challenge
Let us define $\mathcal{P}^n A = \mathcal{P}\mathcal{P} \ldots \mathcal{P}\mathcal{P} A$ where we have applied the power set operation $n$ times.
1. How many elements does $\mathcal{P}^3 \emptyset$ have?
2. How many elements does $\mathcal{P}^n \emptyset$ have? Your answer should depend on $n$.
#### Challenge
Let $X$ be a set. We define the family $(S_x)_{x \in X}$ where $S_x$ is the set of all subsets of $X$ which contain $x$. In other words: [ S_x = \set{ U \subseteq X \mid x \in U } \, . ] Show that
1. $\bigcup_{x \in X} S_x = \mathcal P (X) \setminus { \emptyset}$
2. $\bigcap_{x \in X} S_x = { X }$
#### Challenge
Prove that if $A \subseteq B$ then $\mathcal{P} A \subseteq \mathcal{P} B$. Is the converse true as well? Justify your answer.
## Cartesian Products of Sets
With the tools we have developed we can define the cartesian product $A \times B$ of sets $A$ and $B$ to be the set containing exactly ordered pairs $(a,b)$ where the first component $a$ belongs to the set $A$ and the second component $b$ belongs to $B$. We have expressed the idea of the new object $(a,b)$ but have not defined it so far. Since in set theory, everything is a set we should be able to define $(a,b)$ as a set. Consider the following definition:
$(a, b) \defeq \set{ \set{a}, \set{ a, b } } \in \mathcal{P}(\mathcal{P}(A \cup B)) \,$ where $a \in A$ and $b \in B$.
In other words,
$A \times B \defeq \set{ (a, b) \; \mid a \in A \text{ and } b \in B } \, .$
Notice that if $a = b$, the set $(a, b)$ has only one element:
$(a, a) = \set{ \set{a},\set{a, a} } = \set{ \set{a},\set{a} } = \set{\set{a}} \, .$
#### Challenge
Show that if $a$ and $b$ are distinct then $(a,b) \neq (b,a)$.
The following theorem shows that the definition of cartesian product of sets is reasonable.
#### Theorem
For all elements $a,b,c,d$ of the universe, we have $(a, b) = (c, d)$ if and only if $a = c$ and $b = d$.
#### Challenge
Prove the theorem above.
## Binary Relations
In this lesson our goal is to make the idea of relation and relationship mathematically precise, using the concepts of set theory.
First, consider the following examples of relations:
• The relation on days on the calendar, given by days $x$ and $y$ fall on the same day of the week. For instance, 24 Jan 2022 and 31 Jan 2022 are related with this relationship.
• The relation on vegetable produce in a farmers market, given by price of $x$ is less than price of $y$.
• The relation on people currently alive on the planet, given by $x$ and $y$ have the same home address.
• The relation on people in the world, given by $x$ is a brother of $y$.
• The relation on cities of the world, given by $x$ and $y$ are in the same country.
• The relation on people in the world, given by person $x$ is influenced by person $y$.
• The relation on lines on a 2-dimensional plane, given by line $l$ and line $m$ are parallel to each other.
• The relation on points and lines on a 2-dim plane, given by point $p$ is on line $l$.
• The relation on natural numbers, given by number $m$ is less than $n$.
Now, we will introduce a precise mathematical concept which encompasses all the instance of our informal idea of relation in above. A binary relation $R$ on sets $A$ and $B$ (or from $A$ to $B$) is a two-variable predicate $R$ on the set $A \times B$; Alternatively, we can think of $R$ as a two-variable predicate where the first variable ranges over $A$ and the second over $B$. For every $(a,b) \in A \times B$, the expression $R(a,b)$ is the proposition “$a$ is related to $b$ through $R$”.
In mathematics, we often use infix notation, writing $a \mathrel{R} b$ instead of $R(a, b)$, e.g. $a = b$, $a \leq b$, $f \pitchfork g$, etc.
We call $A$ the domain and $B$ the codomain of the relation $R$.
#### Convention
Whenever the domain and codomain of a relation $R$ are the same set, say $A$, instead of saying $R$ is a relation from $A$ to $A$ we say $R$ is a relation on $A$ (or that $A$ is equipped with a relation $R$).
An extension of a relation $R$ on sets $A$ and $B$ is a subset $[R]$ of $A \times B$ consisting of the pairs $(a,b)$ where $R(a,b)$ holds. Formally,
$[R] \defeq \set{ (a,b) \in A \times B \mid R(a,b)}$
#### Challenge
1. Prove that any subset of $A \times B$ is obtained as an extension of some relation on $A$ and $B$.
2. Use the previous fact to prove that the sets of all relations on $A$ and $B$ coincide with $\mathcal{P} (A \times B)$.
For any set $A$, there there are three trivial relation from $A$ to $A$:
• The maximal relation $\mathrm{m}_A$ which relates every element $a$ of $A$ to every element $a’$ of $A$.
• The identity relation $\mathrm{id}_A$ relates every element of $A$ to itself and nothing else.
• The empty relation $\mathrm{e}_A$ which does not relate any element to any element.
### Inverse of Relations
If we have a relation $R$ from $A$ to $B$ we can form another relation $R^{-1}$, called the inverse of $R$, from $B$ to $A$ defined as follows: $R^{-1}(b,a) \iff R(a,b) \, .$
Note the extension $[R^{-1}]$ is a subset of $B \times A$
### Composition of Relations
Given a relation $R$ from $A$ to $B$ and a relation $S$ from $B$ to $C$ we can compose them to get a a relation $S \circ R$ from $A$ to $C$ defined as follows: $a (S \circ R) c \iff \exists b \in B \, (a R b \land b R c)$
#### Challenge
Draw a picture conveying the idea behind the composition operation of relations.
#### Challenge
Let $B$ be the “brotherhood’’ relation ($x B y$ means $x$ is a brother of $y$) and $S$ be the “sisterhood’’ relation. Show that the composite relation $S \circ B$ is not equivalent to $B$.1
#### Challenge
Show that for any relation $R$ from $A$ to $B$ we always have $R \circ R^{-1} = \mathrm{id}_B$ and $R^{-1} \circ R = \mathrm{id}_A$ as sets.
### Associated Graph of Relations
Suppose a set $A$ comes equipped with a relation $R$. We can associate a directed graph (aka a digraph) whose vertices are elements of $A$ and whose directed edges determined by ordered pairs $(a, b) \in A \times A$ such that $a R b$.
1. Incidentally, I wonder why we use in English “ship” in “relationship”, “kinship”, “friendship”, etc, and “hood” in “sisterhood”, “brotherhood”, “personhood”, “neighbourhood”, etc?Â
## Partially Ordered Sets
A binary relation $R$ on a domain $A$ is a partial order if it has the following three properties:
• Reflexivity: $a R a$, for every $a$ in $A$.
• Transitivity: If $a R b$ and $b R c$, then $a R c$, for every $a$, $b$, and $c$ in $A$.
• Antisymmetry: If $a R b$ and $b R a$ then $a = b$, for every $a$ and $b$ in $A$.
### The Partial Order of Propositions
An natural example of partial order is the partial order on the collection of propositions defined by the relation $P \leq Q$ if $Q$ can be derived from $P$ using the derivation of rules of intuitionistic logic (see here).
#### Challenge
Think why the conditions of reflexivity, transitivity, and antisymmetry hold for this relation.
A partial order $R$ on a domain $A$ is a total order (also called a linear order) if it additionally has the following property: For every $a$ and $b$ in $A$, either $a R b$ holds or $b R a$ holds.
#### Challenge
Express the conditions of reflexivity, transitivity, symmetry, antisymmetry, and totality of order in terms of more familiar connectivity conditions on the associated graph.
### The Partial Order of Propositions
There is a partial order on the power set $\mathcal{P} (X)$ of a set $X$ given by the subset relation:
#### Challenge
1. Check that all the axioms of partial order are satisfied.
2. Show that this partial order is not total.
A non-empty partially ordered set $(S,\leq)$ is filtered (or is said to be a filtered set) if for each $a, b \in S$, there is a element $c$ such that $a \leq c$ and $b \leq c$.
#### Theorem
Every total order is a filtered.
#### Proof
Suppose $A$ is a set equipped with a total order $\leq$. Suppose $a,b \in A$. We want to find an element $c \in A$ for which both $a \leq c$ and $b \leq c$. By totality, either $a \leq b$ or $b \leq a$. If $a \leq b$, then we take $c$ to be $b$, and if $b \leq a$ we take $c$ to be $a$.
#### Challenge
Show that the power set $\mathcal{P}(X)$ with the subset relation is filtered. |
## Calculus with Applications (10th Edition)
Published by Pearson
# Chapter 5 - Graphs and the Derivative - 5.3 Higher Derivatives, Concavity, and the Second Derivative Test - 5.3 Exercises - Page 283: 25
#### Answer
$$f(x)=\ln{x},$$ (a) To find the second derivative of find the first derivative, and then take its derivative. $$f^{\prime}(x)=\frac{1}{x}=x^{-1}=\frac{(-1)^{(0)}. (0)! }{x},$$ $$f^{\prime\prime}(x)=(-1)x^{-2}=-x^{-2} =\frac{(-1)^{(1)}. (1)! }{x^{2}},$$ $$f^{\prime\prime\prime}(x)=-(-2)x^{-3}=2x^{-3}=\frac{(-1)^{(2)}. (2)! }{x^{3}},$$ $$f^{\prime\prime\prime\prime}(x)=2(-3)x^{-4}=-6x^{-4}=\frac{(-1)^{(3)}. (3)! }{x^{4}},,$$ $$f^{(5)}(x)=-6(-4)x^{-5}=24x^{-5}=\frac{(-1)^{(4)}. (4)! }{x^{5}},$$ (b) The $nth$ derivative of $f$ with respect to $x$ is given by $$f^{(n)}(x)=\frac{(-1)^{(n-1)}. (n-1)! }{x^{n}},$$
#### Work Step by Step
$$f(x)=\ln{x},$$ (a) To find the second derivative of find the first derivative, and then take its derivative. $$f^{\prime}(x)=\frac{1}{x}=x^{-1}=\frac{(-1)^{(0)}. (0)! }{x},$$ $$f^{\prime\prime}(x)=(-1)x^{-2}=-x^{-2} =\frac{(-1)^{(1)}. (1)! }{x^{2}},$$ $$f^{\prime\prime\prime}(x)=-(-2)x^{-3}=2x^{-3}=\frac{(-1)^{(2)}. (2)! }{x^{3}},$$ $$f^{\prime\prime\prime\prime}(x)=2(-3)x^{-4}=-6x^{-4}=\frac{(-1)^{(3)}. (3)! }{x^{4}},,$$ $$f^{(5)}(x)=-6(-4)x^{-5}=24x^{-5}=\frac{(-1)^{(4)}. (4)! }{x^{5}},$$ (b) The $nth$ derivative of $f$ with respect to $x$ is given by $$f^{(n)}(x)=\frac{(-1)^{(n-1)}. (n-1)! }{x^{n}},$$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. |
# 2017 IMO Problems/Problem 1
## Problem
For each integer $a_0 > 1$, define the sequence $a_0, a_1, a_2, \ldots$ for $n \geq 0$ as $$a_{n+1} = \begin{cases} \sqrt{a_n} & \text{if } \sqrt{a_n} \text{ is an integer,} \\ a_n + 3 & \text{otherwise.} \end{cases}$$Determine all values of $a_0$ such that there exists a number $A$ such that $a_n = A$ for infinitely many values of $n$.
## Solution
First we observe the following:
When we start with $a_0=3$, we get $a_1=6$, $a_2=9$, $a_3=3$ and the pattern $3,6,9$ repeats.
When we start with $a_0=6$, we get $a_1=9$, $a_2=3$, $a_3=6$ and the pattern $3,6,9$ repeats.
When we start with $a_0=9$, we get $a_1=3$, $a_2=6$, $a_3=9$ and the pattern $3,6,9$ repeats.
When we start with $a_0=12$, we get $a_1=15$, $a_2=15$,..., $a_8=36$, $a_9=6$, $a_{10}=9$, $a_{11}=3$ and the pattern $3,6,9$ repeats.
When this pattern $3,6,9$ repeats, this means that there exists a number $A$ such that $a_n = A$ for infinitely many values of $n$ and that number $A$ is either $3,6,$ or $9$.
When we start with any number $a_0\not\equiv 0\; mod\; 3$, we don't see a repeating pattern.
Therefore the claim is that $a_0=3k$ where $k$ is a positive integer and we need to prove this claim.
When we start with $a_0=3k$, the next term if it is not a square is $3k+3$, then $3k+6$ and so on until we get $3k+3p$ where $p$ is an integer and $(k+p)=3q^2$ where $q$ is an integer. Then the next term will be $\sqrt{9q^2}=3q$ and the pattern repeats again when $q=k$ or when $q=3$ or $6$.
In order for these patterns to repeat, any square in the sequence need to be a multiple of 3.
To try the other two cases where $a_0\not\equiv 0\; mod\; 3$, we can try $a_0=3k\pm 1$ then the next terms will be in the form $3k+3p\pm 1 = 3(k+p) \pm 1$.
When $3(k+p) \pm 1$ is a square, it will not be a multiple of $3$ because $3(k+p) \pm 1$ is not a multiple of $3$ and $3(k+p) \pm 1 \ne 9q^2$ because $3(k+p) \pm 1 \equiv \pm 1\; mod\; 3$ and $q^2$ would have to be $\frac{(k+p)}{3} \pm \frac{1}{9}$ which is not an integer even if $k+p$ is a multiple of $3$.
Therefore the pattern doesn't repeat for any of the other cases where $a_0=3k\pm 1$ and only repeats when $a_0\equiv 0\; mod\; 3$
So, the answer to this problem is $a_0=3k\;\forall k \in \mathbb{Z}^{+}$ and $A=3,6,$ or $9$.
~Tomas Diaz. orders@tomasdiaz.com |
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